This deck is the visual half of College Algebra. It covers plotting and intercepts, the distance and midpoint formulas, and the library of parent-function shapes, then every rigid and non-rigid transformation and the order in which they must be applied, the three symmetry tests, and circles from standard form through completing the square. It targets four classic errors: shifting the wrong way for a horizontal translation, stretching before shifting, mixing up the two reflections, and reading the radius straight off the squared number.
Subject: College Algebra · 138 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
College Algebra - Deck 10
Points and intercepts, the eight parent shapes, how a graph slides, stretches and flips, and circles from the distance formula.
Objectives
This deck turns equations into pictures. Once you can see a graph move, half the rest of the course stops being memorization.
Warm-up
Discussion prompt
Before we open Graphs, Transformations, and Symmetry: without looking back, what was the main idea of Functions, Domain, and Function Notation, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
The single most important idea in College Algebra: a function as a machine with exactly one output per input. Relations versus functions, mapping diagrams and the vertical line test, function notation and evaluating at numbers and expressions, the difference quotient, domain and range in interval notation, piecewise functions, increasing and decreasing intervals, relative extrema, average rate of change, and even versus odd.
Section
Part 1
Concept
Figure (svg): Coordinate plane with the four quadrants labeled I through IV and the point (2, 2) plotted in quadrant I
A point is an ordered pair. The order is the whole point: the first number is the horizontal move, the second is the vertical move.
\[ (x, y) \quad\longrightarrow\quad x \text{ across, then } y \text{ up} \]
quadrant — One of the four regions the axes cut the plane into, numbered counterclockwise starting at the upper right. Points on an axis are in no quadrant at all.
Counterexample
Discussion prompt
A point is an ordered pair. The order is the whole point: the first number is the horizontal move, the second is the vertical move.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Concept
An equation in two variables has infinitely many solutions. Each one is a pair of numbers that makes the equation true.
\[ y = 2x - 1 \quad\Rightarrow\quad (0, -1),\ (1, 1),\ (3, 5),\ \ldots \]
graph of an equation — The set of ALL points whose coordinates make the equation true - and nothing else. A point is on the graph exactly when it satisfies the equation.
Definition probe
Sort into buckets
Every line below is part of the definition of quadrant or of graph of an equation — one or the other, never both. Put each where it belongs.
Picture it
Animation
Shows: An odd function has rotational symmetry — a rendered Manim animation.
Rendered with Manim.
Takeaway: Rotate a half turn about the origin and it lands on itself.
Intuition
Every point in the plane walks up to the equation and asks: do I belong? The equation checks the two coordinates and says yes or no.
The graph is just the crowd of everyone who got a yes. That is why you can always test a point by substitution - no graph paper required.
This is also why a graph never proves anything by itself. The algebra decides; the picture reminds you what the algebra means.
Analogy
Discussion prompt
Explain Think of the graph as a membership test by analogy to something with no College Algebra in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Every point in the plane walks up to the equation and asks: do I belong? The equation checks the two coordinates and says yes or no.
Ranking
Put in order
Put the moves of Graphing by plotting points into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The squared term makes this a parabola, so pick a spread of x-values and let the table tell you where it turns.
Worked example
Sketch the graph by building a table of values.
\[ y = x^2 - 2x - 3 \]
Choose inputs on both sides of where the action is
Why: The squared term makes this a parabola, so pick a spread of x-values and let the table tell you where it turns.
Substitute each x and record the point
Why: Each row is one solution pair. Doing the arithmetic in a column keeps sign errors visible.
| x | value of y | point |
|---|---|---|
| -2 | 5 | (-2, 5) |
| -1 | 0 | (-1, 0) |
| 0 | -3 | (0, -3) |
| 1 | -4 | (1, -4) |
| 2 | -3 | (2, -3) |
| 3 | 0 | (3, 0) |
| 4 | 5 | (4, 5) |
Read the shape off the table before you draw
Why: The y-values fall, bottom out at negative four, then rise - and the table is symmetric about the input one. That is a parabola with its low point at (1, -4).
Verify the point (3, 0) in the original equation
Why: Substituting three gives nine minus six minus three, which is zero. The point checks, and it confirms three is an x-intercept.
\[ (3)^2 - 2(3) - 3 = 9 - 6 - 3 = 0 \quad \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Graphing by plotting points", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Substituting three gives nine minus six minus three, which is zero. The point checks, and it confirms three is an x-intercept.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Graphing this cubic with a lazy table.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: All three chosen inputs happen to be the zeros of the function, so every point landed on the axis.
Sample between the interesting points too.
Why: All three chosen inputs happen to be the zeros of the function, so every point landed on the axis. Three points that agree can agree for a bad reason.
Trap
Graphing this cubic with a lazy table.
\[ y = x^3 - 4x \]
| x | y |
|---|---|
| -2 | 0 |
| 0 | 0 |
| 2 | 0 |
Conclude the graph is the x-axis
Why: All three chosen inputs happen to be the zeros of the function, so every point landed on the axis. Three points that agree can agree for a bad reason.
Sample between the interesting points too.
\[ y = x^3 - 4x \]
| x | y |
|---|---|
| -2 | 0 |
| -1 | 3 |
| 0 | 0 |
| 1 | -3 |
| 2 | 0 |
The curve rises to a hump and drops to a valley
Why: At negative one the value is three, at one it is negative three - the graph is a wave through the three zeros, nothing like the axis. Always add points between your zeros.
Invariant
Step through it
Step through Trap: three points is not a graph one row at a time. One of these columns never changes — find it, and say why it cannot.
Concept
There are only two special places a graph can cross: the horizontal axis and the vertical axis. Each one has a coordinate that is forced to be zero.
| intercept | what is zero there | how to find it |
|---|---|---|
| x-intercept | the y-coordinate | set y to zero, solve for x |
| y-intercept | the x-coordinate | set x to zero, solve for y |
Say it out loud once: to find where it crosses an axis, set the other variable to zero. That sentence prevents the most common intercept mistake.
Comparison
Comparison matrix
From Intercepts are where the graph meets an axis: refill the how to find it column from what you know. The rest of the table is as it appeared.
| intercept | what is zero there | how to find it |
|---|---|---|
| x-intercept | the y-coordinate | set y to zero, solve for x |
| y-intercept | the x-coordinate | set x to zero, solve for y |
Step zero
Discussion prompt
Both intercepts of a line — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: For the x-intercept, set y equal to zero
Answer:
Worked example
Find the x-intercept and the y-intercept.
\[ 3x + 4y = 12 \]
For the x-intercept, set y equal to zero
Why: The x-axis is exactly the set of points whose height is zero, so any crossing point must have y equal to zero.
\[ 3x + 4(0) = 12 \;\Rightarrow\; 3x = 12 \;\Rightarrow\; x = 4 \]
For the y-intercept, set x equal to zero
Why: The y-axis is the set of points whose horizontal position is zero.
\[ 3(0) + 4y = 12 \;\Rightarrow\; 4y = 12 \;\Rightarrow\; y = 3 \]
Write both answers as points, not as bare numbers
Why: An intercept is a location in the plane. A test wants the ordered pair.
\[ (4,\,0) \quad \text{and} \quad (0,\,3) \]
Verify both points in the original equation
Why: Substituting the first pair gives twelve plus zero; substituting the second gives zero plus twelve. Both sides agree in each case.
\[ 3(4) + 4(0) = 12 \;\checkmark \qquad 3(0) + 4(3) = 12 \;\checkmark \]
Picture it
Animation
Shows: Finding both intercepts — a rendered Manim animation.
Rendered with Manim.
Takeaway: The x-intercepts are usually the harder half — that is the solving step.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Hunting the x-intercept of this line.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The name attaches to the wrong variable in your head.
Set the other variable to zero.
Why: The name attaches to the wrong variable in your head. It feels right and it is backwards.
Trap
Hunting the x-intercept of this line.
\[ 2x - 5y = 10 \]
Set x equal to zero because the answer is called the x-intercept
Why: The name attaches to the wrong variable in your head. It feels right and it is backwards.
\[ 2(0) - 5y = 10 \;\Rightarrow\; y = -2 \]
Report the x-intercept as negative two
Why: That number is actually the y-intercept. The reported answer is a real point on the line, which is why the error survives a quick glance.
Set the other variable to zero.
\[ 2x - 5y = 10 \]
For the x-intercept, set y equal to zero
Why: Crossing the x-axis means the height is zero. The name tells you which coordinate you will report, not which one to zero out.
\[ 2x - 5(0) = 10 \;\Rightarrow\; x = 5 \]
Report both, clearly labeled
Why: Checking each: two times five minus zero is ten, and zero minus five times negative two is ten. Both are on the line - they are just different intercepts.
\[ x\text{-intercept } (5, 0) \qquad y\text{-intercept } (0, -2) \]
Notation
Annotate
From Trap: zeroing the variable you are solving for — read this one piece at a time. What is each part doing?
On: \( 2(0) - 5y = 10 \;\Rightarrow\; y = -2 \)
Concept
Figure (svg): Two plotted points joined by a slanted segment, with a horizontal leg and a vertical leg forming a right triangle beneath it
Drop a horizontal leg and a vertical leg from any two points and you have built a right triangle. The segment between them is the hypotenuse.
The horizontal leg is the difference of the first coordinates; the vertical leg is the difference of the second coordinates.
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Because both differences get squared, it never matters which point you call first. Distance cannot come out negative.
Explain it
Discussion prompt
Explain Distance is the Pythagorean theorem in disguise to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
Drop a horizontal leg and a vertical leg from any two points and you have built a right triangle. The segment between them is the hypotenuse.
Estimation
Predict first
Find the distance between the two points.
Commit before you compute: what does Distance between two points come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify by reversing the two points
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Starting from the other point gives differences of negative six and eight; squaring gives thirty-six and sixty-four again, so the distance is still ten.
Worked example
Find the distance between the two points.
\[ (-2,\ 3) \quad \text{and} \quad (4,\ -5) \]
Subtract the first coordinates, then the second coordinates
Why: These two differences are the legs of the right triangle. Keep the subtraction order the same in both slots so you do not scramble the points.
\[ x_2 - x_1 = 4 - (-2) = 6 \qquad y_2 - y_1 = -5 - 3 = -8 \]
Square each leg and add
Why: Squaring erases the negative sign, which is exactly why the order of subtraction cannot hurt you here.
\[ 6^2 + (-8)^2 = 36 + 64 = 100 \]
Take the principal square root
Why: The distance is the positive root, and one hundred is a perfect square, so the answer is exact.
\[ d = \sqrt{100} = 10 \]
Verify by reversing the two points
Why: Starting from the other point gives differences of negative six and eight; squaring gives thirty-six and sixty-four again, so the distance is still ten. The legs six and eight with hypotenuse ten is the familiar right triangle.
\[ \sqrt{(-2-4)^2 + (3-(-5))^2} = \sqrt{36 + 64} = 10 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Distance between two points", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Starting from the other point gives differences of negative six and eight; squaring gives thirty-six and sixty-four again, so the distance is still ten. The legs six and eight with hypotenuse ten is the familiar right triangle.
Concept
The point halfway between two points sits halfway across and halfway up. Halfway between two numbers is their average.
\[ M = \left( \frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2} \right) \]
Notice the plus signs. Distance uses differences; midpoint uses sums. Mixing the two up is the single most common slip on this pair of formulas.
A midpoint is a point, so the answer is an ordered pair. A distance is a length, so the answer is one number.
Worked example
Find the midpoint of the segment joining the two points.
\[ (-2,\ 3) \quad \text{and} \quad (4,\ -5) \]
Average the first coordinates
Why: Halfway between negative two and four on the horizontal axis is the average of the two numbers.
\[ \frac{-2 + 4}{2} = \frac{2}{2} = 1 \]
Average the second coordinates
Why: Same reasoning vertically. Negative five plus three is negative two, and half of that is negative one.
\[ \frac{3 + (-5)}{2} = \frac{-2}{2} = -1 \]
Write the midpoint as an ordered pair
Why: The answer is a location, not a length.
\[ M = (1,\ -1) \]
Verify that the candidate is equidistant from both endpoints
Why: From the first endpoint the legs are three and negative four, giving five; from the second endpoint the legs are three and negative four again, giving five. Five plus five is ten, which is the full distance we already computed.
\[ \sqrt{3^2 + (-4)^2} = 5 \qquad \sqrt{3^2 + (-4)^2} = 5 \qquad 5 + 5 = 10 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Midpoint of the same segment", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: From the first endpoint the legs are three and negative four, giving five; from the second endpoint the legs are three and negative four again, giving five. Five plus five is ten, which is the full distance we already computed.
Ranking
Put in order
These are the steps of Pattern: the point toolkit, scrambled. Put them back in order before the next slide shows you.
Why: This is the order the recipe itself gives. Recalling the sequence without the slide in front of you is the difference between recognising the method and being able to run it — most of what goes wrong in practice is a step done out of turn.
Pattern
Three questions get asked about a pair of points over and over. Here is the whole toolkit.
Memory hook: distance subtracts, midpoint adds. If you wrote a minus sign in a midpoint, you have already made the error.
Sanity check every answer: a midpoint must land between the two endpoints, and a distance must be positive and at least as long as either leg.
Elimination
Eliminate the wrong options
What is the distance between these two points?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: The horizontal difference is negative four minus one, which is negative five; the vertical difference is ten minus negative two, which is twelve. Squaring and adding gives twenty-five plus one hundred forty-four, which is one hundred sixty-nine, and its square root is thirteen.
Check
Work it on paper first - find the two leg lengths before you reach for a calculator.
\[ (1,\ -2) \quad \text{and} \quad (-4,\ 10) \]
Check your understanding
What is the distance between these two points?
Answer: A
Why: The horizontal difference is negative four minus one, which is negative five; the vertical difference is ten minus negative two, which is twelve. Squaring and adding gives twenty-five plus one hundred forty-four, which is one hundred sixty-nine, and its square root is thirteen.
Section
Part 2
Concept
There are not hundreds of graphs to memorize. There are eight basic shapes, and everything else in this course is one of them moved, stretched, or flipped.
parent function — The simplest member of a family of graphs - no shifts, no stretches, no reflections. Every other member of the family is a transformation of it.
For each one you want three things in memory: the shape, the domain and range, and two or three easy points that pin it down.
Picture it
Animation
Shows: The parent functions worth knowing cold — a rendered Manim animation.
Rendered with Manim.
Takeaway: Every graph in this course is one of these, moved.
Intuition
You recognize a friend in a crowd whether they are near or far, sitting or standing. The face is the same; only the position changed.
Parent functions are the faces. Transformations are the position, the size, and the mirror. Once you know the face, a new graph takes seconds instead of a table of twelve points.
So spend real time on this part. Everything after it assumes you can sketch these eight from memory.
Concept
Figure (svg): A horizontal line three units above the x-axis on a coordinate grid
\[ f(x) = c \]
Whatever number you feed it, the same number comes back. Nothing about the output depends on the input, so the graph never rises or falls.
\[ \text{domain } (-\infty, \infty) \qquad \text{range } \{c\} \]
It is symmetric about the vertical axis, and its slope is zero everywhere.
Concept
Figure (svg): A straight line through the origin rising at forty-five degrees
\[ f(x) = x \]
The machine hands your number straight back. Every point on this graph has matching coordinates.
\[ (-3,-3),\quad (0,0),\quad (1,1),\quad (4,4) \]
\[ \text{domain } (-\infty, \infty) \qquad \text{range } (-\infty, \infty) \]
It is symmetric about the origin, which makes it an odd function. Remember this line - inverse graphs get mirrored across it later in the course.
Concept
Figure (svg): A V shape with its corner at the origin, both arms rising at forty-five degrees
\[ f(x) = \left| x \right| \]
Negative inputs come back positive, so the left half of the identity line gets folded up above the horizontal axis.
\[ (-2, 2),\quad (0,0),\quad (2,2) \]
\[ \text{domain } (-\infty, \infty) \qquad \text{range } [0, \infty) \]
The corner at the origin is the giveaway. A parabola turns smoothly; this one turns on a dime.
Concept
Figure (svg): An upward-opening parabola with its vertex at the origin
\[ f(x) = x^2 \]
Squaring kills the sign, so inputs that are opposites give the same output. That is exactly why the two halves match.
\[ (-2, 4),\quad (-1,1),\quad (0,0),\quad (1,1),\quad (2,4) \]
\[ \text{domain } (-\infty, \infty) \qquad \text{range } [0, \infty) \]
Same domain and range as the absolute value shape - the difference is the smooth vertex versus the sharp corner.
Concept
Figure (svg): An S-shaped curve rising through the origin, flat near the origin and steep at both ends
\[ f(x) = x^3 \]
Cubing keeps the sign, so negative inputs give negative outputs. The graph runs from bottom left to top right and never turns around.
\[ (-2,-8),\quad (-1,-1),\quad (0,0),\quad (1,1),\quad (2,8) \]
\[ \text{domain } (-\infty, \infty) \qquad \text{range } (-\infty, \infty) \]
It is symmetric about the origin: spin the picture half a turn and it lands on itself.
Concept
Figure (svg): A curve starting at the origin and rising slowly to the right, with nothing drawn to the left of the origin
\[ f(x) = \sqrt{x} \]
There is no real square root of a negative number, so the graph has a hard left edge. It starts at the origin and there is nothing to its left.
\[ (0,0),\quad (1,1),\quad (4,2),\quad (9,3) \]
\[ \text{domain } [0, \infty) \qquad \text{range } [0, \infty) \]
Use the perfect-square inputs one, four, and nine as your plotting points. They land on whole numbers and save you a calculator.
Concept
Figure (svg): A gentle S-shaped curve through the origin, steep near the origin and flattening at both ends
\[ f(x) = \sqrt[3]{x} \]
An odd root of a negative number is perfectly real, so unlike the square root shape this one keeps going to the left forever.
\[ (-8,-2),\quad (-1,-1),\quad (0,0),\quad (1,1),\quad (8,2) \]
\[ \text{domain } (-\infty, \infty) \qquad \text{range } (-\infty, \infty) \]
It is the mirror image of the cubing shape across the diagonal line - which is a hint about inverses to come.
Concept
Figure (svg): A hyperbola with one branch in the upper right and one in the lower left, hugging both axes
\[ f(x) = \frac{1}{x} \]
Zero is the one input this machine refuses, so the graph splits into two separate pieces that never touch either axis.
\[ \left(-2, -\tfrac{1}{2}\right),\ (-1,-1),\ (1,1),\ \left(2, \tfrac{1}{2}\right) \]
\[ \text{domain } (-\infty, 0) \cup (0, \infty) \qquad \text{range } (-\infty, 0) \cup (0, \infty) \]
Big inputs give tiny outputs and tiny inputs give huge outputs. Both axes act as asymptotes - lines the curve approaches but never reaches.
Concept
Here are all eight side by side. Cover the right three columns and quiz yourself until you can fill them in.
| parent | shape | domain | range |
|---|---|---|---|
| constant | horizontal line | all reals | one number |
| identity | diagonal line | all reals | all reals |
| absolute value | V with a corner | all reals | zero and up |
| squaring | parabola | all reals | zero and up |
| cubing | steep S | all reals | all reals |
| square root | half parabola, right only | zero and up | zero and up |
| cube root | flat S | all reals | all reals |
| reciprocal | two branches | all but zero | all but zero |
Only two of the eight have a restricted domain or range for a structural reason: the square root shape stops at the origin, and the reciprocal skips zero.
Discrimination
Sort into buckets
Sort these by domain, from memory, without looking back at The whole library on one card. Telling them apart on the spot is the skill; the table is only where the answer happens to be written down.
Prediction
Predict first
Which parent function has domain all real numbers, range all numbers greater than or equal to zero, and a sharp corner rather than a smooth turn at the origin?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: the absolute value function
Why: The absolute value shape accepts every real input, never returns a negative output, and meets itself at a sharp corner at the origin. The squaring function matches the domain and range but turns smoothly, so the corner is the deciding clue.
Check
Picture each of the four shapes before you look at the choices.
Check your understanding
Which parent function has domain all real numbers, range all numbers greater than or equal to zero, and a sharp corner rather than a smooth turn at the origin?
Answer: A
Why: The absolute value shape accepts every real input, never returns a negative output, and meets itself at a sharp corner at the origin. The squaring function matches the domain and range but turns smoothly, so the corner is the deciding clue.
Section
Part 3
Concept
A function is a machine. There are exactly two places you can tamper with it: before the number goes in, or after the answer comes out.
\[ y = a\, f\big(b(x - h)\big) + k \]
| where the change is | what it does | which direction |
|---|---|---|
| outside, added | shift | vertical |
| outside, multiplied | stretch, compress, or flip | vertical |
| inside, added | shift | horizontal |
| inside, multiplied | stretch, compress, or flip | horizontal |
That table is the whole part. Everything that follows is just filling in the details of those four rows.
Discrimination
Sort into buckets
Sort these by what it does, from memory, without looking back at Every transformation touches the input or the…. Telling them apart on the spot is the skill; the table is only where the answer happens to be written down.
Picture it
Animation
Shows: Several transformations at once — a rendered Manim animation.
Rendered with Manim.
Takeaway: Flipped, squashed, shifted up, and sliding sideways.
Intuition
Changes on the outside happen to the answer after the function is done, so they do exactly what they look like. Add three and the graph goes up three.
Changes on the inside happen to the input before the function runs. They do the opposite of what they look like, and every student is surprised by it at least once.
Here is why, in one sentence: if you add three to the input first, the machine reaches its old answer sooner, so the picture arrives earlier - which is to the left.
Concept
\[ y = f(x) + k \]
Every output is nudged by the same amount, so the entire picture rides up or down without changing shape at all.
| equation | direction | point that was on the graph | point that is now on it |
|---|---|---|---|
| y equals f(x) plus 4 | up 4 units | (2, 5) | (2, 9) |
| y equals f(x) minus 4 | down 4 units | (2, 5) | (2, 1) |
The first coordinate never changes for a vertical shift. Only the height moves.
Trade off
Comparison matrix
From Vertical shift: add outside, move up or down: every row here is a choice with a cost. Fill the point that was on the graph column, then say which row you would actually pick and what you give up for it.
| equation | direction | point that was on the graph | point that is now on it |
|---|---|---|---|
| y equals f(x) plus 4 | up 4 units | (2, 5) | (2, 9) |
| y equals f(x) minus 4 | down 4 units | (2, 5) | (2, 1) |
Step zero
Discussion prompt
Graphing a vertical shift — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Name the parent and the change
Answer:
Worked example
Graph this function by transforming a parent shape, not by guessing points.
\[ g(x) = \left| x \right| - 2 \]
Name the parent and the change
Why: Inside the bars there is nothing but the variable, so no horizontal move. The two is subtracted outside, which is a vertical shift down two.
Move each key point of the V down two units
Why: A vertical shift keeps the first coordinate and lowers the second, so the corner slides from the origin straight down.
| point on the parent | point on the new graph |
|---|---|
| (-2, 2) | (-2, 0) |
| (0, 0) | (0, -2) |
| (2, 2) | (2, 0) |
Read off the new features
Why: The corner is now two units below the origin, and the arms cross the horizontal axis at two places instead of touching it once.
\[ \text{corner } (0,-2) \qquad \text{range } [-2, \infty) \]
Verify the point (2, 0) in the original rule
Why: The absolute value of two is two, and two minus two is zero, so the point really is on the graph. That confirms the shift, not just the picture.
\[ g(2) = \left| 2 \right| - 2 = 2 - 2 = 0 \quad \checkmark \]
Picture it
Animation
Shows: Adding outside shifts vertically — a rendered Manim animation.
Rendered with Manim.
Takeaway: Outside the function, and the graph moves the way you expect.
Concept
Figure (svg): A dashed parabola with its vertex at the origin and a solid identical parabola whose vertex sits three units to the right, with an arrow between the two vertices
\[ y = f(x - h) \]
Subtracting a number inside moves the graph right. Adding a number inside moves it left. Yes, it is backwards from what it looks like.
| equation | direction | point that was on the graph | point that is now on it |
|---|---|---|---|
| y equals f of the quantity x minus 3 | right 3 units | (2, 5) | (5, 5) |
| y equals f of the quantity x plus 3 | left 3 units | (2, 5) | (-1, 5) |
The second coordinate never changes for a horizontal shift. Only the position left or right moves.
Intuition
Do not memorize the flip. Derive it in two seconds with one question: what input makes the inside equal the old input?
\[ y = f(x - 3) \;\text{ hits } f(0) \text{ when } x - 3 = 0,\ \text{ that is } x = 3 \]
The value the parent produced at zero now happens at three. The graph did not decide to go right - the bookkeeping forced it.
Another way to say it: the inside expression is a toll you pay before entering. Subtracting three means you have to travel three further to pay it.
Hypothesis
Predict first
Graphing a horizontal shift is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Name the parent and the change
Why: The square root shape is the parent. The four is subtracted inside the radical, so this is a horizontal shift, and subtracting inside moves the graph right.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
Graph this function and state its domain.
\[ g(x) = \sqrt{x - 4} \]
Name the parent and the change
Why: The square root shape is the parent. The four is subtracted inside the radical, so this is a horizontal shift, and subtracting inside moves the graph right.
Find where the graph now starts
Why: The parent starts where the inside is zero. Setting the inside to zero gives the starting input, and that point carries the parent's starting height of zero.
\[ x - 4 = 0 \;\Rightarrow\; x = 4 \quad\text{so the graph starts at } (4, 0) \]
Slide each perfect-square point right four
Why: Every parent point keeps its height and moves four units right, so the easy whole-number points stay easy.
| point on the parent | point on the new graph |
|---|---|
| (0, 0) | (4, 0) |
| (1, 1) | (5, 1) |
| (4, 2) | (8, 2) |
| (9, 3) | (13, 3) |
State the domain from the shifted starting point
Why: The expression under the radical must not be negative, and it is exactly zero at four, so the graph lives from four rightward.
\[ \text{domain } [4, \infty) \]
Verify the point (8, 2) in the original rule
Why: Eight minus four is four, and the principal square root of four is two, so the point sits on the graph exactly as the shift predicted.
\[ g(8) = \sqrt{8 - 4} = \sqrt{4} = 2 \quad \checkmark \]
Picture it
Animation
Shows: Adding inside shifts the other way — a rendered Manim animation.
Rendered with Manim.
Takeaway: Inside the bracket, and the motion is opposite to the sign.
Trap
Locating the vertex of this parabola.
\[ y = (x - 3)^2 \]
See the minus sign and shift left three
Why: Minus means left everywhere else in life, so the hand writes the vertex at negative three without asking.
\[ \text{claimed vertex } (-3, 0) \]
Test the claim and watch it fail
Why: Substituting negative three gives negative six squared, which is thirty-six - nowhere near zero. The claimed vertex is not even close to the bottom of the graph.
\[ y = (-3 - 3)^2 = (-6)^2 = 36 \neq 0 \]
Ask what input makes the inside zero.
\[ y = (x - 3)^2 \]
Solve the inside equal to zero
Why: The parent parabola bottoms out when its input is zero, so the new graph bottoms out when the inside expression is zero.
\[ x - 3 = 0 \;\Rightarrow\; x = 3 \quad\text{vertex } (3, 0) \]
Confirm with a substitution
Why: Substituting three gives zero squared, which is zero - the lowest possible output of a square. The graph moved right three, opposite the sign you see.
\[ y = (3-3)^2 = 0 \quad \checkmark \]
Break the constraint
Discussion prompt
The rule this trap just fixed:
The parent parabola bottoms out when its input is zero, so the new graph bottoms out when the inside expression is zero.
Now break it on purpose. Build a case that violates it and follow the consequences until something visibly fails. Where does the failure first show up — and would you have noticed it if you had not been looking?
Hint: The dangerous rules are the ones whose violation still produces an answer. If yours fails loudly, try to find one that fails quietly.
Answer:
Minus means left everywhere else in life, so the hand writes the vertex at negative three without asking.
Concept
\[ y = a\, f(x), \qquad a > 0 \]
Multiplying the output scales every height. A factor bigger than one makes the graph taller and narrower; a factor between zero and one makes it shorter and wider.
| factor | effect | the point (2, 4) becomes |
|---|---|---|
| 3 | vertical stretch by 3 | (2, 12) |
| one half | vertical compression by one half | (2, 2) |
| 1 | no change | (2, 4) |
Points on the horizontal axis do not move at all - zero times anything is still zero. That is why a stretch looks like the graph is being pulled away from that axis.
Comparison
Comparison matrix
From Vertical stretch and compression: refill the effect column from what you know. The rest of the table is as it appeared.
| factor | effect | the point (2, 4) becomes |
|---|---|---|
| 3 | vertical stretch by 3 | (2, 12) |
| one half | vertical compression by one half | (2, 2) |
| 1 | no change | (2, 4) |
Picture it
Animation
Shows: Multiplying outside stretches — a rendered Manim animation.
Rendered with Manim.
Takeaway: Bigger than one stretches; between zero and one squashes.
Ranking
Put in order
Put the moves of Stretching and compressing a parabola into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Every other column is a multiple of this one, so getting it right once protects both of the others.
Worked example
Compare the three graphs by building one table.
\[ y = x^2, \qquad y = 2x^2, \qquad y = \tfrac{1}{2}x^2 \]
Compute the parent column first
Why: Every other column is a multiple of this one, so getting it right once protects both of the others.
Multiply that column by two, then by one half
Why: A vertical scaling touches only the output, so the inputs across a row never change.
| x | x squared | 2 times x squared | half of x squared |
|---|---|---|---|
| -2 | 4 | 8 | 2 |
| -1 | 1 | 2 | 0.5 |
| 0 | 0 | 0 | 0 |
| 1 | 1 | 2 | 0.5 |
| 2 | 4 | 8 | 2 |
Describe what you see
Why: At the same input of two, the stretched graph is already at eight while the compressed one is only at two. Taller means narrower-looking, because it reaches a given height sooner.
Verify the stretched value at the input two
Why: Two squared is four, and twice four is eight, which matches the table entry exactly. The shared point at the origin also checks, since twice zero is zero.
\[ 2(2)^2 = 2 \cdot 4 = 8 \quad \checkmark \]
Concept
\[ y = f(bx), \qquad b > 0 \]
This is an inside change, so it is backwards again. Multiplying the input by a number bigger than one squeezes the graph toward the vertical axis.
| equation | effect | the point (6, 5) becomes |
|---|---|---|
| y equals f of 2x | horizontal compression by one half | (3, 5) |
| y equals f of half of x | horizontal stretch by 2 | (12, 5) |
Check the first row the reliable way: at three, the inside doubles to six, which is exactly the input the parent needed. So the height five now happens at three.
The heights never change here - only how far out you have to go to reach them.
Picture it
Animation
Shows: Multiplying inside compresses — a rendered Manim animation.
Rendered with Manim.
Takeaway: Inside again, and again the effect is the reciprocal of what you expect.
Concept
Figure (svg): The square root curve drawn solid above the x-axis, its mirror image dashed below the x-axis, and a second dashed mirror image to the left of the y-axis
\[ y = -f(x) \]
The minus sign sits outside, so it changes the answer. Every output flips sign and the whole graph folds across the horizontal axis like a card.
\[ (4, 2) \;\longrightarrow\; (4, -2) \]
The domain is untouched. The range flips upside down.
Concept
\[ y = f(-x) \]
Now the minus sign is inside, so it changes the input. Every point swings to the other side of the vertical axis, keeping its height.
\[ (4, 2) \;\longrightarrow\; (-4, 2) \]
This time the range is untouched and the domain flips. For the square root shape that is dramatic: the graph moves from the right half of the plane to the left half.
Picture it
Animation
Shows: A negative flips it — a rendered Manim animation.
Rendered with Manim.
Takeaway: Passing through zero turns the parabola upside down.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Sketching the flipped square root graph.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The word negative gets attached to the horizontal direction, so the sketch lands in the second quadrant.
Ask first: is the minus inside or outside?
Why: The word negative gets attached to the horizontal direction, so the sketch lands in the second quadrant.
Trap
Sketching the flipped square root graph.
\[ y = -\sqrt{x} \]
Read the minus as flipping the graph to the left side
Why: The word negative gets attached to the horizontal direction, so the sketch lands in the second quadrant.
\[ \text{claimed point } (-4, 2) \]
Test the claim at negative four
Why: There is no real square root of negative four, so the rule does not even produce a number there. The claimed point is not on the graph, and neither is anything else to the left.
\[ -\sqrt{-4} \;\text{ is not a real number} \]
Ask first: is the minus inside or outside?
\[ y = -\sqrt{x} \qquad \text{versus} \qquad y = \sqrt{-x} \]
Outside minus flips the outputs downward
Why: The input four still gives a square root of two, and the minus then makes it negative. The graph stays on the right and hangs below the axis.
\[ y = -\sqrt{4} = -2 \;\Rightarrow\; (4, -2) \]
Inside minus flips the inputs sideways
Why: Now negative four is legal, because the minus makes the inside positive four, whose root is two. This is the graph that lives on the left.
\[ y = \sqrt{-(-4)} = \sqrt{4} = 2 \;\Rightarrow\; (-4, 2) \;\checkmark \]
Translation
\( -\sqrt{-4} \;\text{ is not a real number} \)
Draw it
Translate both ways. First write the expression above as a sentence with no symbols in it at all. Then cover it, and write your sentence back as notation. If the two versions disagree, the disagreement is the thing to fix.
Concept
With one transformation there is nothing to decide. With two or more, doing them in the wrong order gives a genuinely different graph.
The rule behind the list is just order of operations on a single input: whatever happens to the number first is the transformation you apply first.
The one you must never get backwards is the last pair: stretch first, shift second.
Picture it
Animation
Shows: Order matters — a rendered Manim animation.
Rendered with Manim.
Takeaway: Shift-then-stretch and stretch-then-shift give different graphs.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Building the graph described as: shift the parabola up three, then stretch it vertically by two.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Doing the moves in the order the sentence lists them feels obedient, but the doubling now lands on the three as well.
Read the equation and follow the order of operations.
Why: Doing the moves in the order the sentence lists them feels obedient, but the doubling now lands on the three as well.
Trap
Building the graph described as: shift the parabola up three, then stretch it vertically by two.
\[ \text{start from } y = x^2 \]
Shift up three first, then double everything
Why: Doing the moves in the order the sentence lists them feels obedient, but the doubling now lands on the three as well.
\[ y = 2\left(x^2 + 3\right) = 2x^2 + 6 \]
Read the resulting vertex
Why: Substituting zero gives six, so this graph sits six units up. That is not the graph anyone meant to write when they wrote a plus three at the end.
\[ \text{vertex } (0, 6) \]
Read the equation and follow the order of operations.
\[ y = 2x^2 + 3 \]
Stretch first, then shift
Why: Given an input, you square it, then double it, and only then add three. The addition is last, so the shift is applied last.
\[ x \;\to\; x^2 \;\to\; 2x^2 \;\to\; 2x^2 + 3 \]
Read the resulting vertex and compare
Why: Substituting zero gives three, not six. The two orders differ by a full three units of height, so this is not a technicality.
\[ \text{vertex } (0,3) \quad \checkmark \]
Notation
Annotate
From Trap: shifting before stretching — read this one piece at a time. What is each part doing?
On: \( \text{vertex } (0,3) \quad \checkmark \)
Step zero
Discussion prompt
All four at once — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: List the transformations in order
Answer:
Worked example
Describe the transformations, then find four points and the domain and range.
\[ g(x) = -2\sqrt{x + 3} + 1 \]
List the transformations in order
Why: Inside first: plus three means left three. Then the outside factor of negative two is a vertical stretch by two together with a flip across the horizontal axis. The plus one is the vertical shift, applied last.
Write the rule that moves a single point
Why: Applying the same four moves to a parent point in order gives one formula you can run on every point in the table.
\[ (x,\, y) \;\longrightarrow\; (x - 3,\ -2y + 1) \]
Run the four easy parent points through it
Why: The perfect-square inputs keep the arithmetic exact, so every new point is a whole-number pair you can plot with confidence.
| parent point | new first coordinate | new second coordinate | new point |
|---|---|---|---|
| (0, 0) | -3 | 1 | (-3, 1) |
| (1, 1) | -2 | -1 | (-2, -1) |
| (4, 2) | 1 | -3 | (1, -3) |
| (9, 3) | 6 | -5 | (6, -5) |
State the domain and range
Why: The radical starts at negative three and runs right forever. The outputs start at one and fall forever, because the negative factor turns the rising parent into a falling graph.
\[ \text{domain } [-3, \infty) \qquad \text{range } (-\infty, 1] \]
Verify the point (1, -3) in the original rule
Why: One plus three is four, whose root is two; negative two times two is negative four; and negative four plus one is negative three. Every step of the transformation order shows up in that computation.
\[ g(1) = -2\sqrt{1+3} + 1 = -2(2) + 1 = -3 \quad \checkmark \]
Pattern
\[ y = a\, f\big(b(x-h)\big) + k \]
Step two is the one people skip. If the inside is written as two x minus six, the shift is right three, not right six, because the inside factors as two times the quantity x minus three.
Picture it
Animation
Shows: Reading a graph for its features — a rendered Manim animation.
Rendered with Manim.
Takeaway: Intercepts, turning points, and end behaviour — in that order.
Estimation
Predict first
Start with the absolute value parent. Reflect it across the horizontal axis, stretch it vertically by four, then shift it left three and up one. Write the equation.
Commit before you compute: what does Writing the equation from a description come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify the corner and one arm point
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. At negative three the expression gives one, matching the corner.
Worked example
Start with the absolute value parent. Reflect it across the horizontal axis, stretch it vertically by four, then shift it left three and up one. Write the equation.
Start with the parent and place the horizontal shift inside
Why: Left three means the graph arrives three units early, so the inside must be the variable plus three - opposite the direction named.
\[ y = \left| x + 3 \right| \]
Attach the stretch and the flip as one outside factor
Why: A vertical stretch by four multiplies the outputs by four, and a reflection across the horizontal axis makes them negative. One coefficient of negative four does both.
\[ y = -4\left| x + 3 \right| \]
Add the vertical shift last
Why: The vertical shift is applied after the scaling, so it goes on the very outside as a plus one.
\[ y = -4\left| x + 3 \right| + 1 \]
Locate the corner
Why: The absolute value is zero when its inside is zero, which happens at negative three, and there the output is just the one.
\[ x + 3 = 0 \;\Rightarrow\; x = -3 \quad\text{corner } (-3, 1) \]
Verify the corner and one arm point
Why: At negative three the expression gives one, matching the corner. At negative two the inside is one, four times one is four, the flip makes it negative four, and adding one gives negative three - so the graph does fall away from the corner as a reflected V should.
\[ y(-3) = -4(0)+1 = 1 \;\checkmark \qquad y(-2) = -4(1)+1 = -3 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Writing the equation from a description", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At negative three the expression gives one, matching the corner. At negative two the inside is one, four times one is four, the flip makes it negative four, and adding one gives negative three - so the graph does fall away from the corner as a reflected V should.
Check
Decide which number is inside and which is outside before you pick.
\[ y = \sqrt{x + 4} - 2 \]
Check your understanding
How is this graph obtained from the square root parent function?
Answer: A
Why: The four is added inside the radical, and inside changes run opposite to their sign, so the graph moves left four; the two is subtracted outside, which lowers every output by two. The starting point moves from the origin to the point negative four, negative two.
Commit first
Predict first
Which equation describes the transformed graph?
Commit to an answer, then rate it — certain, fairly sure, or guessing — and write the rating down before you turn the page.
Correct: y equals negative three times the quantity x minus one, squared, plus four
Why: Right one puts a minus one inside the parentheses, the stretch by three and the flip combine into an outside coefficient of negative three, and up four is added last. Checking the vertex: at one the squared term is zero, leaving four, so the vertex is the point one, four.
The rating matters as much as the answer: confident-and-wrong is the combination that survives revision, because nothing about it feels like it needs revisiting.
Check
Write it yourself first, then look for your answer among the choices.
Take the squaring parent. Stretch it vertically by a factor of three, reflect it across the horizontal axis, then shift it right one and up four.
Check your understanding
Which equation describes the transformed graph?
Answer: A
Why: Right one puts a minus one inside the parentheses, the stretch by three and the flip combine into an outside coefficient of negative three, and up four is added last. Checking the vertex: at one the squared term is zero, leaving four, so the vertex is the point one, four.
Section
Part 4
Intuition
Symmetry is a physical question you can ask about a drawing, and each kind of symmetry is a different physical move.
Each physical move has an algebraic twin: folding across an axis flips the sign of one coordinate, and spinning flips the sign of both.
That is the entire idea. The tests you are about to learn are those sign flips written down.
Concept
Folding across the vertical axis sends a point to its mirror image with the same height and the opposite horizontal position.
\[ (x, y) \;\longrightarrow\; (-x,\, y) \]
So the test is: swap the sign of the first variable everywhere and see whether the equation comes back unchanged.
\[ f(-x) = f(x) \quad \text{for every } x \]
even function — A function whose graph is symmetric about the vertical axis. Only even powers of the variable survive the sign flip, which is where the name comes from.
Matching
Match the pairs
Match each term to the definition this lesson gave it — not the one you would guess from the word.
Why: These are the working definitions of quadrant, graph of an equation, parent function, even function as Graphs, Transformations, and Symmetry uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Concept
Spinning the page half a turn sends a point to the spot diagonally opposite through the origin - both coordinates change sign.
\[ (x, y) \;\longrightarrow\; (-x,\, -y) \]
\[ f(-x) = -f(x) \quad \text{for every } x \]
odd function — A function whose graph is symmetric about the origin. Flipping the input's sign flips the whole output's sign. The cubing, identity, cube root, and reciprocal parents are all odd.
Most functions are neither even nor odd. Failing both tests is the normal outcome, not a mistake.
Concept
The third fold sends a point to the same horizontal position at the opposite height.
\[ (x, y) \;\longrightarrow\; (x,\, -y) \]
This one is different in an important way: a graph with it fails the vertical line test, because the same input has two different outputs.
\[ x = y^2 \quad\text{contains both } (4, 2) \text{ and } (4, -2) \]
So no function except the flat zero function can have this symmetry. It shows up for equations solved for the first variable, and for circles.
Explain it
Discussion prompt
Explain Symmetry about the horizontal axis to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
The third fold sends a point to the same horizontal position at the opposite height.
Picture it
Animation
Shows: Testing for symmetry algebraically — a rendered Manim animation.
Rendered with Manim.
Takeaway: Substitute minus x and compare — no graph required.
Pattern
Take the equation as given, make the substitution, then simplify and compare with the original. Nothing else.
| symmetry | substitute | symmetric when |
|---|---|---|
| vertical axis | replace the first variable with its negative | the simplified equation matches the original |
| horizontal axis | replace the second variable with its negative | the simplified equation matches the original |
| origin | replace both variables with their negatives | the simplified equation matches the original |
Two shortcuts once you trust them: a polynomial with only even powers is symmetric about the vertical axis, and a polynomial with only odd powers and no constant term is symmetric about the origin.
A single mixed term ruins both. Testing takes ten seconds - do it rather than guessing from the shortcut.
Trade off
Comparison matrix
From Pattern: the three symmetry tests: every row here is a choice with a cost. Fill the symmetric when column, then say which row you would actually pick and what you give up for it.
| symmetry | substitute | symmetric when |
|---|---|---|
| vertical axis | replace the first variable with its negative | the simplified equation matches the original |
| horizontal axis | replace the second variable with its negative | the simplified equation matches the original |
| origin | replace both variables with their negatives | the simplified equation matches the original |
Estimation
Predict first
Test this equation for all three symmetries.
Commit before you compute: what does Running all three tests on one equation come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify with a mirrored pair of points
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. At one the value is one minus three, which is negative two; at negative one the value is one minus three again, still negative two.
Worked example
Test this equation for all three symmetries.
\[ y = x^4 - 3x^2 \]
Test the vertical axis by negating the first variable
Why: An even power of a negative is positive, so the fourth power and the square both come back exactly as they were. The equation is unchanged, so this symmetry holds.
\[ (-x)^4 - 3(-x)^2 = x^4 - 3x^2 \quad \checkmark \]
Test the horizontal axis by negating the second variable
Why: The left side becomes negative while the right side is untouched, so solving for the second variable flips every sign on the right. That is a different equation.
\[ -y = x^4 - 3x^2 \;\Rightarrow\; y = -x^4 + 3x^2 \quad \text{not the same} \]
Test the origin by negating both variables
Why: The right side is unchanged by the first flip, so the whole test collapses to the horizontal-axis test we just failed. No origin symmetry.
\[ -y = (-x)^4 - 3(-x)^2 = x^4 - 3x^2 \quad \text{not the same} \]
State the conclusion
Why: Exactly one test passed, so this is an even function and nothing more.
\[ \text{symmetric about the vertical axis only} \]
Verify with a mirrored pair of points
Why: At one the value is one minus three, which is negative two; at negative one the value is one minus three again, still negative two. Equal heights at opposite inputs is precisely what the vertical-axis fold means, and since negative two is not equal to two, the horizontal-axis symmetry really does fail.
\[ y(1) = 1 - 3 = -2 \qquad y(-1) = 1 - 3 = -2 \quad \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Running all three tests on one equation", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At one the value is one minus three, which is negative two; at negative one the value is one minus three again, still negative two. Equal heights at opposite inputs is precisely what the vertical-axis fold means, and since negative two is not equal to two, the horizontal-axis symmetry really does fail.
Prediction
Predict first
Which equation has a graph that is symmetric with respect to the origin?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: y equals x cubed minus four x
Why: Negating the input gives the negative of the cube plus four times the input, which is exactly the negative of the whole original expression, so negating both variables returns the original equation. Both powers present are odd and there is no constant term.
Check
Negate both variables in each choice before you decide.
Check your understanding
Which equation has a graph that is symmetric with respect to the origin?
Answer: A
Why: Negating the input gives the negative of the cube plus four times the input, which is exactly the negative of the whole original expression, so negating both variables returns the original equation. Both powers present are odd and there is no constant term.
Section
Part 5
Fill the middle
Fill in the blanks
From A circle is a distance condition — finish the line. Write what belongs on the right of the equals sign before you look.
(x - h)^2 + (y - k)^2 = r^2
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Both sides are non-negative, so squaring loses nothing, and it turns the awkward radical into the clean standard form.
Concept
Figure (svg): A circle centered at the point two, negative one with a marked center and a radius segment of length three drawn to the right
A circle is not really a new object. It is the set of all points whose distance from one fixed point is a fixed number.
\[ \sqrt{(x - h)^2 + (y - k)^2} = r \]
Square both sides to clear the radical
Why: Both sides are non-negative, so squaring loses nothing, and it turns the awkward radical into the clean standard form.
\[ (x - h)^2 + (y - k)^2 = r^2 \]
That is the whole formula, and it is just the distance formula with the radical squared away. The center is the pair of numbers being subtracted, and the number on the right is the radius squared.
Analogy
Discussion prompt
Explain A circle is a distance condition by analogy to something with no College Algebra in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
A circle is not really a new object. It is the set of all points whose distance from one fixed point is a fixed number.
Concept
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Two habits keep this painless. First, the coordinates of the center are the opposites of the numbers you see, exactly like a horizontal shift.
| written equation | center | radius |
|---|---|---|
| (x minus 5) squared plus (y minus 2) squared equals 9 | (5, 2) | 3 |
| (x plus 5) squared plus (y minus 2) squared equals 9 | (-5, 2) | 3 |
| x squared plus (y plus 1) squared equals 16 | (0, -1) | 4 |
Second, the number on the right is never the radius. Take its square root, every single time.
Comparison
Comparison matrix
From Reading a circle out of standard form: refill the radius column from what you know. The rest of the table is as it appeared.
| written equation | center | radius |
|---|---|---|
| (x minus 5) squared plus (y minus 2) squared equals 9 | (5, 2) | 3 |
| (x plus 5) squared plus (y minus 2) squared equals 9 | (-5, 2) | 3 |
| x squared plus (y plus 1) squared equals 16 | (0, -1) | 4 |
Step zero
Discussion prompt
Writing a circle from its center and radius — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Subtract each center coordinate inside its own square
Answer:
Worked example
Write the standard-form equation of the circle with the given center and radius.
\[ \text{center } (-3,\ 4), \qquad r = 6 \]
Subtract each center coordinate inside its own square
Why: The formula subtracts the center from the point, so a center coordinate of negative three becomes the variable minus a negative three.
\[ \big(x - (-3)\big)^2 + (y - 4)^2 = r^2 \]
Simplify the double negative
Why: Subtracting a negative is adding, so a center to the left of the vertical axis shows up as a plus sign inside the parentheses.
\[ (x + 3)^2 + (y - 4)^2 = r^2 \]
Square the radius on the right
Why: Standard form stores the square of the radius, not the radius itself, because that is what squaring the distance formula produced.
\[ (x + 3)^2 + (y - 4)^2 = 36 \]
Verify with a point six units from the center
Why: Starting at the center and moving six units right lands on the point three, four, which must be on the circle. Substituting gives six squared plus zero, which is thirty-six, matching the right side exactly.
\[ (3+3)^2 + (4-4)^2 = 36 + 0 = 36 \quad \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Writing a circle from its center and radius", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Starting at the center and moving six units right lands on the point three, four, which must be on the circle. Substituting gives six squared plus zero, which is thirty-six, matching the right side exactly.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Call the radius twenty-five and step that far from the center
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The number is sitting right there on the right side, so the hand copies it.
The right side is the radius squared.
Why: The number is sitting right there on the right side, so the hand copies it. Nothing in the equation announces that it has been squared.
Trap
Graphing this circle.
\[ (x - 1)^2 + (y + 2)^2 = 25 \]
Call the radius twenty-five and step that far from the center
Why: The number is sitting right there on the right side, so the hand copies it. Nothing in the equation announces that it has been squared.
\[ \text{claimed point } (26,\ -2) \]
Test the claimed point
Why: Twenty-six minus one is twenty-five, and twenty-five squared is six hundred twenty-five - not twenty-five. The point is nowhere near the circle, and the sketch would be twenty-five times too wide.
\[ (26-1)^2 + (-2+2)^2 = 625 \neq 25 \]
The right side is the radius squared.
\[ (x - 1)^2 + (y + 2)^2 = 25 \]
Take the square root of the right side
Why: Standard form came from squaring the distance formula, so undoing that square is the last step of reading the equation.
\[ r = \sqrt{25} = 5 \qquad \text{center } (1,\ -2) \]
Step five units from the center and confirm
Why: Moving five right of the center gives the point six, negative two. Substituting gives five squared plus zero, which is twenty-five - it lands on the circle.
\[ (6-1)^2 + (-2+2)^2 = 25 \quad \checkmark \]
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
Concept
If you multiply a circle's standard form out and move everything to one side, the center and radius vanish from view.
\[ x^2 + y^2 + Dx + Ey + F = 0 \]
You can still recognize it: both variables are squared, the two squared terms have the same coefficient, and there is no term with both variables multiplied together.
To graph it you must get back to standard form, and the tool for that is completing the square - once for each variable.
Counterexample
Discussion prompt
If you multiply a circle's standard form out and move everything to one side, the center and radius vanish from view.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
You can still recognize it: both variables are squared, the two squared terms have the same coefficient, and there is no term with both variables multiplied together.
Picture it
Animation
Shows: Transformations move the domain and range — a rendered Manim animation.
Rendered with Manim.
Takeaway: Horizontal shifts move the domain; vertical ones move the range.
Ranking
Put in order
Put the moves of General form to standard form into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Completing the square works on one variable at a time, so the two families have to be separated first, and the loose constant has to be out of the way.
Worked example
Find the center and radius, then sketch.
\[ x^2 + y^2 - 6x + 8y - 11 = 0 \]
Group the terms of each variable and move the constant right
Why: Completing the square works on one variable at a time, so the two families have to be separated first, and the loose constant has to be out of the way.
\[ \left(x^2 - 6x\right) + \left(y^2 + 8y\right) = 11 \]
Halve each linear coefficient and square it
Why: Half of negative six is negative three, and its square is nine. Half of eight is four, and its square is sixteen. Those are the numbers that turn each group into a perfect square.
\[ \left(\tfrac{-6}{2}\right)^2 = 9 \qquad \left(\tfrac{8}{2}\right)^2 = 16 \]
Add both numbers to both sides
Why: Adding only on the left would change the equation. Nine and sixteen go onto the right as well, and eleven plus nine plus sixteen is thirty-six.
\[ \left(x^2 - 6x + 9\right) + \left(y^2 + 8y + 16\right) = 11 + 9 + 16 = 36 \]
Write each group as a square
Why: Each perfect-square trinomial factors as the variable plus half of its linear coefficient, squared - negative three for the first, positive four for the second.
\[ (x - 3)^2 + (y + 4)^2 = 36 \]
Read the center and radius
Why: The center coordinates are the opposites of the numbers inside, and the radius is the square root of the right side, not the right side itself.
\[ \text{center } (3,\ -4) \qquad r = \sqrt{36} = 6 \]
Verify by expanding back to the original equation
Why: The first square expands to the square term minus six times the variable plus nine, the second to the square term plus eight times the variable plus sixteen. Their sum is thirty-six, and moving everything to one side gives the constant twenty-five minus thirty-six, which is negative eleven - the original equation exactly.
\[ x^2 - 6x + 9 + y^2 + 8y + 16 = 36 \;\Rightarrow\; x^2 + y^2 - 6x + 8y - 11 = 0 \quad \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "General form to standard form", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The first square expands to the square term minus six times the variable plus nine, the second to the square term plus eight times the variable plus sixteen. Their sum is thirty-six, and moving everything to one side gives the constant twenty-five minus thirty-six, which is negative eleven - the original equation exactly.
Constraint
Discussion prompt
Run Pattern: any circle, start to finish with this step confiscated:
Read the radius as the square root of the right-hand side.
Is it still possible? If it is, say what takes its place and what it costs you. If it is not, say exactly what that step was providing that nothing else does.
Hint: A step you can drop for free was never load-bearing. If you cannot drop it, name the thing that goes wrong the moment it is gone.
Answer:
Pattern
Two quick sanity checks: a negative number on the right means there are no points at all, and a right side of zero means the whole circle collapses to the single center point.
Edge cases
Discussion prompt
Pattern: any circle, start to finish works on the cases you have just seen. Push it to the edge: what is the most degenerate input it still handles — empty, zero, one item, everything equal — and what is the first case where it stops being true? Name the case, not just "it breaks".
Hint: Try the smallest legal input, then the largest, then the one where two things collide. Methods are specified at their edges; the middle takes care of itself.
Answer:
Two quick sanity checks: a negative number on the right means there are no points at all, and a right side of zero means the whole circle collapses to the single center point.
Elimination
Eliminate the wrong options
What are the center and radius of this circle?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: Standard form subtracts the center, so a written plus three means the second coordinate of the center is negative three, and the missing first group means the first coordinate is zero. The right side is the radius squared, and the square root of forty-nine is seven.
Check
Watch both the sign inside and the number on the right.
\[ x^2 + (y + 3)^2 = 49 \]
Check your understanding
What are the center and radius of this circle?
Answer: A
Why: Standard form subtracts the center, so a written plus three means the second coordinate of the center is negative three, and the missing first group means the first coordinate is zero. The right side is the radius squared, and the square root of forty-nine is seven.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — Points, Intercepts, Distance · The Library of Parent Shapes · Moving the Graph · Symmetry · Circles. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
You started this deck with equations and you are leaving it with pictures you can draw without a table of values.
| if you see | it means | the classic error |
|---|---|---|
| a number added inside | horizontal shift, opposite direction | shifting the way the sign looks |
| a number multiplied outside | vertical stretch or compression | shifting before stretching |
| a minus sign outside | flip across the horizontal axis | flipping sideways instead |
| a number on the right of a circle | the radius squared | using it as the radius |
Next up: linear functions and slope, where every one of these moves gets a rate-of-change meaning.
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