Sketching the graph of a quadratic function. Includes the standard form of a quadratic function, the parabola and which way it opens, the vertex and the axis of symmetry, the formula for the vertex's x-coordinate, building a table of values around the vertex, and reading the y-intercept straight off the equation.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 9 — Quadratic Equations and Functions
Graphing Quadratic Functions
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.4 Graphing Quadratic Functions §9.4, pp. 520-525 — the lesson these objectives are drawn from
Warm-up
Chapter 4 graphed straight lines from tables. The same method works here, with one extra step at the start.
Discussion prompt
Make a table of values for y equal to x squared, using x from negative three to three. What do you notice about the y values?
Hint: Compare the value at three with the value at negative three.
Answer:
\[ x: -3, -2, -1, 0, 1, 2, 3 \quad \longrightarrow \quad y: 9, 4, 1, 0, 1, 4, 9 \]
The values read the same forwards and backwards, because squaring destroys the sign. That symmetry is the single most useful fact about these graphs, and it halves the work of drawing every one of them.
Concept
A quadratic function is one that can be written as y equals a times x squared plus b times x plus c, with a not nought. Every such function has a U-shaped graph called a parabola.
parabola — The U-shaped graph of a quadratic function. It opens upwards when the leading coefficient is positive and downwards when it is negative.
The sign of a alone decides which way it opens.
Figure (svg): Two parabolas, one opening up and one opening down
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.4 Graphing Quadratic Functions §9.4, pp. 520-520
Section
Section 1
Concept
The graph of a quadratic function opens up when a is positive and down when a is negative. An upward parabola has a lowest point and a downward one has a highest point.
\[ y = ax^2 + bx + c, \quad a \ne 0 \]
Only the sign of a matters, not b or c.
Figure (svg): Two parabolas, one opening up and one opening down
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.4 Graphing Quadratic Functions §9.4, pp. 520-520 — Example 1, Describe the Graph of a Parabola
Picture it
Lowest point or highest point.
Figure (svg): Two parabolas, one opening up and one opening down
The two curves have the same width because their leading coefficients have the same size. Only the sign differs, and only the direction changes.
Worked example
This is Example 1 from the textbook.
\[ \text{Describe the graphs of } y = x^2 \text{ and } y = -x^2 + 4. \]
Read the first leading coefficient
Why: The coefficient of x squared is one.
\[ a = 1,\text{ positive} \]
Say which way it opens
Why: Positive a opens up.
\[ \text{lowest point } (0, 0) \]
Read the second
Why: The coefficient is negative one.
\[ a = -1,\text{ negative} \]
Say which way it opens
Why: Negative a opens down.
\[ \text{highest point } (0, 4) \]
Figure (svg): Two parabolas, one opening up and one opening down
\[ y = x^2 \text{ opens up}; \quad y = -x^2 + 4 \text{ opens down} \]
Verify: test a large value of x in each
Why: At x equal to three the first gives nine and the second gives negative five. The first climbs away from its lowest point and the second falls away from its highest, which is what opening up and opening down mean.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.4 Graphing Quadratic Functions §9.4, pp. 520-520
Sorting
Read only the coefficient of x squared.
Sort into buckets
Sort each function by which way its parabola opens.
The last one is written with its terms reversed, which is exactly the case where the coefficient is easiest to misread. Rewriting it as negative x squared plus four settles it.
Worked example
Guided Practice 1 to 3, without drawing anything.
\[ \text{Does each open up or down: } y = x^2, \; y = -2x^2 + 4, \; y = 3x^2 + 5x - 1? \]
Take the first
Why: The coefficient of x squared is one.
Take the second
Why: The coefficient is negative two.
Take the third
Why: The coefficient is three.
Note what was ignored
Why: The other terms never came into it.
Figure (svg): The solution to Worked example three more, by inspection shown as a ladder of expressions, one row per algebraic move
\[ \text{up}, \quad \text{down}, \quad \text{up} \]
Verify: check the third against a far-out value
Why: At x equal to ten the third gives three hundred and forty-nine, a large positive number, and at negative ten it gives two hundred and forty-nine. Both ends rise, which is what opening up looks like from a distance.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.4 Graphing Quadratic Functions §9.4, pp. 520-520
Trap
\[ y = -x^2 + 4 \]
The four is positive, so the parabola opens up
Why: The most visible number was the one that got read.
The four moves the curve up the page but does not turn it over. The coefficient of x squared is negative one, so the parabola opens down and four is its highest point rather than its lowest.
\[ a = -1 < 0 \;\Longrightarrow\; \text{opens down} \]
Read the coefficient of the squared term and no other
Why: That is the only number the direction depends on.
Rewriting the function with the squared term first makes the coefficient easier to find.
Faded example
The squared term only.
Fill in the blanks
In y = -2x squared + 4, the leading coefficient is -2, so the parabola opens down.
Why: The minus sign belongs to the coefficient, not just to the term, and it is what turns the curve over. The four raises the whole graph and leaves its direction alone.
Elimination
Check for a squared term with a non-zero coefficient.
Eliminate the wrong options
Which of these is not a quadratic function?
Survives elimination: A
Why: The definition requires a squared term whose coefficient is not nought, and the first has no squared term at all — it is linear, and its graph is a straight line. Missing b or c terms are perfectly allowed.
Socratic
The other coefficients do not.
Discussion prompt
Explain why a negative leading coefficient makes the parabola open downwards. Then say what happens to the graph for very large values of x, either positive or negative.
Hint: Ask which term dominates when x is large.
Answer:
For large x the squared term grows far faster than the linear and constant terms, so it eventually controls the sign of the whole expression. Since x squared is positive at both ends, multiplying it by a positive a sends the graph upwards on both sides and multiplying by a negative a sends it downwards on both sides.
At x equal to a hundred, the squared term of a function like negative x squared minus three x plus one contributes negative ten thousand while the other terms contribute less than three hundred. The far behaviour of every quadratic is decided by its leading term alone, which is why b and c can shift and tilt the picture without ever flipping it.
Section
Section 2
Concept
The vertex is the highest or lowest point on a parabola. The vertical line through the vertex divides the curve into two mirror-image halves and is called the axis of symmetry.
axis of symmetry — The vertical line through the vertex of a parabola, dividing it into two symmetric halves. For y equals a x squared plus b x plus c it is the line x equals negative b over two a.
Folding along it makes the halves match exactly.
Figure (svg): A parabola with its vertex and axis of symmetry labelled
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.4 Graphing Quadratic Functions §9.4, pp. 521-521 — the definitions of vertex and axis of symmetry and the Study Tip on folding the graph
Picture it
The fold that makes them match.
Figure (svg): A parabola with its vertex and axis of symmetry labelled
Every point on the curve has a partner the same distance from the axis on the other side, and both have the same y value. That is what a table around the vertex makes visible.
Worked example
Using the formula on Example 2's function.
\[ \text{Find the vertex and axis of symmetry of } y = x^2 - 2x - 3. \]
Identify the coefficients
Why: Read them off the standard form.
\[ a = 1, b = -2, c = -3 \]
Apply the formula
Why: Negative b over two a.
\[ x = 2 \div 2 = 1 \]
Find the y-coordinate
Why: Substitute one into the function.
\[ 1 - 2 - 3 = -4 \]
State the axis
Why: The vertical line through the vertex.
\[ x = 1 \]
Figure (svg): A parabola with its vertex and axis of symmetry labelled
\[ \text{vertex } (1, -4), \quad \text{axis } x = 1 \]
Verify: test the symmetry with a pair of points
Why: At x equal to nought the value is negative three, and at x equal to two it is also negative three. Those two points are each one unit from the axis, so the symmetry holds exactly where it should.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.4 Graphing Quadratic Functions §9.4, pp. 521-521
Faded example
Mind the sign of b.
Fill in the blanks
y = x^2 - 2x - 3: \quad x = \dfrac-2})}1 = ___
Why: Writing b as negative two before negating it makes the double sign change explicit. Reading b as two instead would put the vertex at negative one, on the wrong side of the y-axis.
Worked example
Example 3's function, where the signs need care.
\[ \text{Find the vertex and axis of } y = -x^2 - 3x + 1. \]
Identify the coefficients
Why: Both a and b are negative.
\[ a = -1, b = -3, c = 1 \]
Apply the formula
Why: Negative of negative three, over negative two.
\[ x = 3 \div (-2) \]
Simplify
Why: Negative three halves.
\[ x = -1\tfrac{1}{2} \]
Find the y-coordinate
Why: Substitute it back.
\[ y = 3\tfrac{1}{4} \]
Figure (svg): A parabola whose vertex has a fractional x-coordinate
\[ \text{vertex } \left(-\tfrac{3}{2}, \tfrac{13}{4}\right), \quad \text{axis } x = -\tfrac{3}{2} \]
Verify: check the symmetry across the fraction
Why: At x equal to negative one the value is three, and at negative two it is also three. Both are half a unit from the axis at negative one and a half, so the fractional vertex is exactly where the symmetry says it should be.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.4 Graphing Quadratic Functions §9.4, pp. 522-522
Trap
\[ y = x^2 - 2x - 3 \;\Longrightarrow\; x = \dfrac{-2}{2(1)} = -1 \]
Put the two from the middle term into the formula
Why: The coefficient looked like two.
The coefficient is negative two, and the formula asks for its negative, so the numerator is positive two and the vertex is at one. The wrong answer puts the vertex on the wrong side of the axis entirely.
\[ b = -2 \;\Longrightarrow\; x = \dfrac{-(-2)}{2(1)} = 1 \]
Write b with its own sign, then negate it
Why: Two sign changes are easier to track than one done in your head.
Checking a pair of symmetric points afterwards catches this in one line.
Matching
Negative b over two a.
Match the pairs
Why: Two of these have no middle term, so b is nought and the axis is the y-axis itself. A missing x term always means the parabola is centred on the y-axis, which is worth recognising instantly.
Prediction
Functions like y equals x squared plus four.
Predict first
Where is the axis of symmetry when b is nought?
Correct: The y-axis, since negative b over two a is zero.
\[ b = 0 \;\Longrightarrow\; x = \dfrac{-0}{2a} = 0 \]
Why: With b equal to nought the formula gives nought regardless of what a and c are, so the axis is the line x equals nought. That is why y equals x squared and y equals negative x squared plus four are both centred on the y-axis despite looking quite different — c raises or lowers the curve and a flips or stretches it, but neither shifts it sideways.
Socratic
Not every graph has one.
Discussion prompt
Explain where the symmetry of a parabola comes from. Then say why the axis is always vertical rather than slanted.
Hint: Think about what squaring does to a distance.
Answer:
The function depends on x through a squared term, and squaring gives the same result for a quantity and its negative. Two x values the same distance either side of the vertex differ from it by amounts that are negatives of each other, so the squared part is identical for both and the two points have the same height.
The axis is vertical because the pairing is between x values, with y determined by them — the symmetry swaps left for right and never top for bottom. A slanted or horizontal axis would require a relation that is not a function of x in the ordinary way, which is exactly the kind of curve Lesson 12.1 begins to look at.
Section
Section 3
Concept
To sketch a parabola, find the x-coordinate of the vertex, make a table using x values to the left and right of it, plot the points, and connect them with a smooth curve.
Three points either side is usually plenty.
Figure (svg): A table of values showing the symmetry about the vertex
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.4 Graphing Quadratic Functions §9.4, pp. 521-521 — the Graphing a Quadratic Function steps and Example 2
Picture it
Symmetric values either side.
Figure (svg): A table of values showing the symmetry about the vertex
The repetition in the second row is a free check on the arithmetic. If the values are not symmetric about the vertex column, either the vertex or one of the entries is wrong.
Worked example
This is Example 2 from the textbook.
\[ \text{Sketch the graph of } y = x^2 - 2x - 3. \]
Find the vertex's x-coordinate
Why: Negative b over two a.
\[ x = 1 \]
Build the table
Why: Three values either side.
\[ x \text{ from } -2 \text{ to } 4 \]
Compute the values
Why: Symmetric about the vertex.
\[ 5, 0, -3, -4, -3, 0, 5 \]
Plot and connect
Why: A smooth curve opening up.
\[ \text{vertex } (1, -4) \]
Figure (svg): A parabola with its vertex and axis of symmetry labelled
\[ \text{vertex } (1, -4), \quad \text{axis } x = 1 \]
Verify: check the two x-intercepts
Why: The table gives y equal to nought at x equal to negative one and at x equal to three, and those two points are each two units from the axis at x equal to one. Crossing points are always symmetric about the axis, which is the fact Lesson 9.5 turns into a method.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.4 Graphing Quadratic Functions §9.4, pp. 521-521
Faded example
The vertex decides the range.
Fill in the blanks
\text-2 x = 1 \;\Longrightarrow\; \text4 x = ___ \text___ x = ___
Why: Equal reach on both sides makes the symmetry visible in the values and the sketch balanced. Starting at nought out of habit would hide the left branch entirely.
Worked example
The same function, computing only half the table.
\[ \text{Complete the table for } y = x^2 - 2x - 3 \text{ from the left half alone.} \]
Compute to the left of the vertex
Why: x equal to negative two, negative one, nought.
\[ 5, 0, -3 \]
Compute the vertex
Why: x equal to one.
\[ -4 \]
Reflect the values
Why: Each x value has a mirror partner.
\[ -3, 0, 5 \]
Name the partners
Why: Nought pairs with two, and so on.
\[ 0\leftrightarrow 2, \; -1\leftrightarrow 3 \]
Figure (svg): A table of values showing the symmetry about the vertex
\[ 5, \; 0, \; -3, \; -4, \; -3, \; 0, \; 5 \]
Verify: spot-check one reflected value
Why: At x equal to four the function gives sixteen minus eight minus three, which is five — the same as at negative two, as the reflection predicted. Using symmetry halves the arithmetic and computing one reflected value confirms it was applied correctly.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.4 Graphing Quadratic Functions §9.4, pp. 521-521
Error analysis
The student graphed y equals x squared minus two x minus three using a table from negative three to three.
Annotate
On: \( \begin{aligned} x: &\; -3, \; -2, \; -1, \; 0, \; 1, \; 2, \; 3 \\ y: &\; 12, \; 5, \; 0, \; -3, \; -4, \; -3, \; 0 \end{aligned} \)
Nothing here is arithmetically wrong, which is what makes it worth studying. The table was built out of habit around nought rather than around the vertex, and the whole point of finding the vertex first is to decide where the table should sit.
Sorting
Match the action to what it gives you.
Sort into buckets
Sort each action by what it is for.
Four of the six steps are about finding points and two are about the shape they should form. Knowing the shape first is what tells you the joined curve should be smooth rather than a series of corners.
Elimination
For a parabola with vertex at x equal to three.
Eliminate the wrong options
Which range of x values should the table use?
Survives elimination: A
Why: Reaching three units either side of the vertex shows both branches and makes the symmetry visible. Each wrong option produces a sketch that looks like half a curve or a curve with no turning point at all.
Socratic
The plotted points could be joined with segments.
Discussion prompt
Explain why a parabola is drawn as a smooth curve rather than by connecting the dots with straight lines. Then say what the segments would get wrong near the vertex.
Hint: Ask what the function does between the plotted x values.
Answer:
The function has a value at every x, not only at the whole numbers in the table, and those intermediate values do not lie on the straight lines between plotted points. A quadratic bends continuously, so a segmented drawing is only an approximation that happens to be exact at the seven points chosen.
Near the vertex the error is worst, because that is where the curve bends most sharply. A segment from the point at x equal to nought to the point at x equal to one would pass below the actual curve for the whole interval, and joining the two lowest points with a straight line would even suggest a flat bottom rather than a single turning point.
Section
Section 4
Concept
When a is negative the parabola opens down and the vertex is the highest point. When the vertex's x-coordinate is a fraction, the table can still use whole numbers either side of it.
Only the vertex itself needs the fraction.
Figure (svg): A parabola whose vertex has a fractional x-coordinate
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.4 Graphing Quadratic Functions §9.4, pp. 522-522 — Example 3 and its Study Tip on choosing whole numbers for the table
Picture it
Whole numbers still work.
Figure (svg): A parabola whose vertex has a fractional x-coordinate
The values at negative one and negative two are equal, which is the symmetry announcing that the vertex sits halfway between them. Equal values in a table always bracket the axis.
Worked example
This is Example 3 from the textbook.
\[ \text{Sketch the graph of } y = -x^2 - 3x + 1. \]
Find the vertex's x-coordinate
Why: Three over negative two.
\[ x = -1\tfrac{1}{2} \]
Build the table
Why: Whole numbers either side, plus the vertex.
\[ -4 \text{ to } 1 \]
Compute the values
Why: Symmetric about negative one and a half.
\[ -3, 1, 3, 3\tfrac{1}{4}, 3, 1, -3 \]
Plot and connect
Why: Opens down since a is negative.
\[ vertex \left(-1\tfrac{1}{2}, 3\tfrac{1}{4}\right) \]
Figure (svg): A parabola whose vertex has a fractional x-coordinate
\[ \text{vertex } \left(-\tfrac{3}{2}, \tfrac{13}{4}\right), \quad \text{axis } x = -\tfrac{3}{2} \]
Verify: check that the vertex is the highest value
Why: Every value in the table is at most three and a quarter, and that maximum occurs at the vertex. For a downward parabola the vertex value should be the largest in the table, and if some other entry exceeded it the vertex would be in the wrong place.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.4 Graphing Quadratic Functions §9.4, pp. 522-522
Faded example
Both a and b are negative here.
Fill in the blanks
a = -1, \; b = -3: \quad x = \dfrac-2-1.5 = \dfrac______} = ___
Why: The numerator becomes positive because b was negative, and the denominator is negative because a is. Two negatives in different places is the sign pattern that catches most people out.
Worked example
Reading the symmetry backwards.
\[ \text{A table gives equal values at } x = -1 \text{ and } x = -2. \text{ Where is the axis?} \]
Recall the symmetry
Why: Equal values sit at equal distances from the axis.
Find the midpoint
Why: Halfway between negative one and negative two.
\[ -1\tfrac{1}{2} \]
State the axis
Why: The vertical line through it.
\[ x = -1\tfrac{1}{2} \]
Cross-check with the formula
Why: Negative b over two a agrees.
Figure (svg): A parabola whose vertex has a fractional x-coordinate
\[ x = \dfrac{-1 + (-2)}{2} = -\tfrac{3}{2} \]
Verify: confirm against the formula
Why: For this function b is negative three and a is negative one, so negative b over two a is three over negative two, which is negative one and a half. Two independent routes to the same number is as good a check as this lesson offers.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.4 Graphing Quadratic Functions §9.4, pp. 522-522
Trap
\[ x = -\tfrac{3}{2} \approx -2 \]
Round the vertex to the nearest whole number for convenience
Why: The table uses whole numbers anyway.
The vertex is a specific point on the curve, and at negative two the value is three rather than the maximum of three and a quarter. Labelling the wrong point as the vertex misstates the highest value the function reaches.
\[ \text{vertex } \left(-\tfrac{3}{2}, \tfrac{13}{4}\right) \]
Keep the vertex exact and choose whole numbers only for the other table entries
Why: The two choices are independent.
The Study Tip says exactly this: a fractional vertex does not force fractional table values.
Hypothesis
Two different x values give the same y.
Predict first
If y is the same at x equal to negative one and at x equal to negative two, where is the axis?
Correct: Exactly halfway between them, at -1.5.
\[ \dfrac{-1 + (-2)}{2} = -\tfrac{3}{2} \]
Why: Points with the same height are mirror images across the axis, so the axis must be the same distance from each — which puts it at their midpoint. This works for any such pair, not just adjacent ones: equal values at nought and at four would place the axis at two. It is often faster than the formula when a table is already in front of you.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| a positive | a negative | |
|---|---|---|
| Opens | up | down |
| The vertex is | the lowest point | the highest point |
| In a table, the vertex value is | the smallest | the largest |
The last row is a useful check on any table: for an upward parabola no entry should fall below the vertex value. If one does, the vertex was computed wrongly.
Socratic
The table still uses whole numbers.
Discussion prompt
Explain why a vertex with a fractional x-coordinate does not force the table to use fractions. Then say what would be lost by leaving the vertex out of the sketch entirely.
Hint: Ask what the table is for.
Answer:
The table exists to supply plottable points on the curve, and any x values will do for that — they do not have to include the vertex or be spaced around it evenly. Whole numbers either side of a fractional vertex give perfectly good points, and the symmetry still shows up as equal values in pairs.
Leaving the vertex out would mean the sketch has no point at the turning point itself, so the drawn curve would have to guess where the maximum or minimum sits and how high it reaches. Since the vertex is usually the most interesting feature — the greatest height, the least cost, the peak of a throw — plotting it explicitly is worth the one extra substitution.
Section
Section 5
Concept
Standard form gives up three features immediately: the direction from the sign of a, the axis and vertex from negative b over two a, and the y-intercept from c.
That is enough for a rough sketch with no table at all.
Figure (svg): The constant term read off as the y-intercept
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.4 Graphing Quadratic Functions §9.4, pp. 522-522 — the Summary box, Graph of a Quadratic Function, and the note that the y-intercept is c
Picture it
Set x to nought.
Figure (svg): The constant term read off as the y-intercept
The constant term is a plotted point that costs nothing to find. Adding it to a table is a good habit even when the table already covers x equal to nought.
Worked example
Using only the summary's three facts.
\[ \text{Describe the graph of } y = -x^2 - 3x + 1 \text{ without making a table.} \]
Read the direction
Why: The leading coefficient is negative.
Find the axis
Why: Negative b over two a.
\[ x = -1\tfrac{1}{2} \]
Find the vertex height
Why: Substitute the axis value.
\[ y = 3\tfrac{1}{4} \]
Read the y-intercept
Why: The constant term.
\[ y = 1 \]
Figure (svg): The constant term read off as the y-intercept
\[ \text{down}; \; \left(-\tfrac{3}{2}, \tfrac{13}{4}\right); \; (0, 1) \]
Verify: check the y-intercept by substitution
Why: Setting x to nought gives negative nought minus nought plus one, which is one, matching the constant term. That will always happen, which is why the constant can be read off rather than computed.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.4 Graphing Quadratic Functions §9.4, pp. 522-522
Translation
The constant term.
Match the pairs
Why: The last one has no constant written, which means c is nought and the curve passes through the origin. A missing term always means a coefficient of nought rather than a missing feature.
Worked example
Three facts and the symmetry.
\[ \text{Sketch } y = x^2 - 2x - 3 \text{ from its features and one extra point.} \]
Note direction and vertex
Why: Opens up, turning at one.
\[ (1, -4) \]
Plot the y-intercept
Why: The constant term.
\[ (0, -3) \]
Reflect it across the axis
Why: One unit each side of x equal to one.
\[ (2, -3) \]
Draw the curve
Why: Through three points, opening up.
Figure (svg): A parabola with its vertex and axis of symmetry labelled
\[ (0, -3), \; (1, -4), \; (2, -3) \]
Verify: check the reflected point directly
Why: At x equal to two the function gives four minus four minus three, which is negative three, exactly as the reflection predicted. Using the axis to mirror the y-intercept produces a third point with no arithmetic at all.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.4 Graphing Quadratic Functions §9.4, pp. 521-522
Trap
\[ y = x^2 - 2x - 3 \;\Longrightarrow\; \text{lowest point } -3 \]
The constant is negative three, so that is the lowest value
Why: The constant is the y-intercept and it looked like the bottom.
Negative three is the height where the curve crosses the y-axis, not its minimum. The vertex is at x equal to one, where the value is negative four — a full unit lower.
\[ \text{y-intercept } -3; \quad \text{vertex } (1, -4) \]
Keep the two facts separate: c gives a point, the formula gives the vertex
Why: They coincide only when b is nought.
When b is nought the axis is the y-axis and the two really are the same point, which is where the confusion starts.
Faded example
Substitute nought for x.
Fill in the blanks
y = a(0)^2 + b(0) + c = c \quad \textc ___
Why: Both variable terms carry a factor of x, so both become nought and only the constant survives. That is why the y-intercept never requires any work.
Elimination
Given a quadratic in standard form.
Eliminate the wrong options
Which of these cannot simply be read off the equation?
Survives elimination: A
Why: The vertex's height requires computing the axis and then substituting it back, which is two steps rather than a reading. Everything else in the list is visible in the equation as written, which is what makes standard form worth recognising.
Socratic
Three features and the symmetry.
Discussion prompt
Describe how to sketch a parabola from its direction, vertex and y-intercept alone. Then say when a full table would still be worth making.
Hint: The y-intercept has a mirror partner.
Answer:
The vertex fixes the turning point, the direction says which way the arms go, and the y-intercept gives a third point — which can be reflected across the axis for a fourth at no cost. Four points and a known shape are enough for a sketch that shows the position, direction and rough width of the curve.
A table is worth making when the width matters, when the graph is being used to read off values, or when the vertex is close to the y-intercept so the two points are nearly on top of each other and give little information about the arms. For a quick description or a rough picture the three features are usually enough, which is why the summary lists them together.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Feature | Where it comes from | For y = x squared - 2x - 3 |
|---|---|---|
| Direction | the sign of a | opens up |
| Axis of symmetry | x = -b over 2a | x = 1 |
| y-intercept | the constant term c | -3 |
Two of the three rows need no arithmetic at all. Only the axis requires a calculation, and even that is a single division.
Pattern
To sketch the graph of any quadratic function, these five moves cover it.
Step four is where the table gets centred, and centring it on the vertex rather than on nought is what makes the sketch balanced and the symmetry visible.
OpenStax Elementary Algebra 2e, §10.5 Graphing Quadratic Equations in Two Variables §10.5
Check
The leading coefficient.
Check your understanding
Which way does the graph of y = -2x squared + 4 open?
Answer: A
Why: The coefficient of the squared term is negative two, which is negative, so the parabola opens down with a highest point at four.
Check
Mind the sign of b.
Check your understanding
What is the axis of symmetry of y = x squared - 2x - 3?
Answer: A
Why: With a equal to one and b equal to negative two, negative b over two a is two over two, which is one.
Check
Set x to nought.
Check your understanding
What is the y-intercept of y = -x squared - 3x + 1?
Answer: A
Why: The y-intercept is the constant term, because setting x to nought makes both variable terms vanish.
Real world
This is the shot put question from the lesson opener. The path of a thrown shot is a parabola, and its highest point is the vertex.
Discussion prompt
Suppose a throw follows the path given by height equals negative sixteen t squared plus thirty-two t plus six, in feet, with t in seconds. Find when the shot is highest, how high it goes, and what the constant term tells you.
Hint: The highest point is the vertex.
Answer:
\[ t = \dfrac{-32}{2(-16)} = 1 \;\Longrightarrow\; h = -16 + 32 + 6 = 22 \]
The shot reaches its highest point one second after release, at twenty-two feet, and the parabola opens down because the leading coefficient is negative — as it must, since the shot comes back down.
The constant term of six is the y-intercept, which here is the height at the moment of release: the thrower's hand was six feet above the ground. Every one of the three features read straight off the equation has a physical meaning in this situation, which is what makes standard form worth recognising rather than merely memorising.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
For y equals x squared minus two x minus three, which point is the vertex?
Correct: (1, -4).
\[ x = \dfrac{-(-2)}{2(1)} = 1, \quad y = 1 - 2 - 3 = -4 \]
Why: The axis of symmetry is at negative b over two a, which is two over two, or one, and substituting one gives one minus two minus three, which is negative four. The y-intercept at negative three is a point on the curve but not its lowest — it sits one unit above the vertex, and its mirror partner at x equal to two has the same height, which is a quick way to see that neither can be the turning point. The crossing point at negative one is where the curve meets the axis, not where it turns. The vertex and the y-intercept coincide only when b is nought, and here b is negative two.
Explain it
They built their table from negative three to three out of habit and got a lopsided sketch.
Discussion prompt
In no more than four sentences, explain why the table should be centred on the vertex. Then tell them how to know where to start it.
Hint: The vertex comes before the table.
Answer:
A usable answer: a parabola is symmetric about its vertex, so a table centred anywhere else shows more of one branch than the other and the sketch looks unbalanced. Centring it on the vertex makes the values repeat in pairs, which both looks right and checks your arithmetic.
To know where to start, compute negative b over two a first and then take about three whole numbers each side of it. That is why finding the vertex is step one rather than step two.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: The sign is fixed by writing b with its own sign first and negating it as a separate step. The direction is fixed by reading only the coefficient of the squared term. Centring is fixed by finding the vertex before touching the table. The last one is fixed by remembering that c gives a point on the y-axis while the vertex is at negative b over two a, and the two coincide only when b is nought. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of a page write the standard form of a quadratic function and label a, b and c, then write beside each one what it tells you about the graph. Underneath, draw two axes side by side and sketch an upward and a downward parabola, marking the vertex on each and labelling it as the lowest or highest point. In the middle, graph y equals x squared minus two x minus three completely: compute the axis, build a table from negative two to four, write the y values in a row so the repetition is visible, plot the points and draw the axis of symmetry as a dashed line. Beneath that, do the same for y equals negative x squared minus three x plus one, taking care with the two negative signs in the vertex formula and keeping the vertex as a fraction while the table uses whole numbers. In the lower corner, write the three features that can be read from the equation with no table at all. Finally, in the margin, write why equal values in a table locate the axis halfway between them.
Your two tables should each read the same forwards and backwards about the vertex column. If one does not, either the vertex is misplaced or one entry was computed wrongly, and the symmetry tells you which entry to recheck.
Recap
Five things, and the first three need no plotting at all.
| If the question says | Your first move is |
|---|---|
| Which way does it open? | Read the sign of a alone |
| Find the vertex | Compute -b over 2a, then substitute back |
| Find the axis of symmetry | The vertical line x = -b over 2a |
| Sketch the graph | Find the vertex, then centre a table on it |
| Find the y-intercept | Read the constant term |
Lesson 9.5 uses these graphs to solve equations. The points where a parabola crosses the x-axis are exactly the solutions of the corresponding quadratic equation — which finally explains why such an equation can have two solutions, one, or none.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.4 Graphing Quadratic Functions §9.4, pp. 520-525 — everything on these slides traces back here
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