The square root and cube root parent functions with their domains and ranges, vertical stretches, shrinks and reflections, translating a square root function and reading its new domain and range, translating a cube root function, and a pendulum period model.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 6 — Rational Exponents and Radical Functions
Graph Square Root and Cube Root Functions
Objectives
Five outcomes. The parity of the index runs through all of them.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.5 Graph Square Root and Cube Root Functions §6.5, pp. 446-451 — the lesson these objectives are drawn from
Warm-up
Lesson 6.4 produced the square root and cube root as inverses. This lesson draws them.
Discussion prompt
The square root function is the inverse of the squaring function restricted to non-negative inputs, and the cube root is the inverse of cubing with no restriction. What does each of those facts tell you about the two graphs before you plot a single point?
Hint: The domain of an inverse is the range of the original.
Answer:
The squaring function on the non-negatives has domain and range both non-negative, so the square root's domain and range are both x at least zero. Its graph is that parabola reflected in the line y equals x — half a parabola lying on its side.
The cubing function has domain and range all real numbers, so the cube root's do too. Its graph is the cubic reflected in the same line, unbroken and running away in both directions.
Concept
A radical function is built from one of two parent graphs by a stretch or shrink, possibly a reflection, and a translation. Which parent is used decides the shape and the domain; the numbers a, h and k decide everything else.
radical function — A function whose rule contains a variable inside a radical. The two families in this lesson are a times the square root of x minus h, plus k, and the same with a cube root.
\[ y = a\sqrt{x-h}+k, \qquad y = a\sqrt[3]{x-h}+k \]
The letters do exactly what they did for the absolute value graphs of Lesson 2.7 and the parabolas of Lesson 4.2: a for stretch and direction, h for horizontal shift, k for vertical shift.
Figure (svg): Two columns contrasting the shape and domain of a square root graph with a cube root graph
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.5 Graph Square Root and Cube Root Functions §6.5, pp. 446-448
Section
Section 1
Concept
The square root parent function has domain and range both restricted to the non-negatives, so its graph starts at the origin and goes up and to the right. The cube root parent accepts every real number and produces every real number, so its graph runs unbroken through the origin.
parent function — The simplest function of a family, from which every other member is obtained by stretching, reflecting and translating. Here the parents are the square root of x and the cube root of x.
\[ f(x) = \sqrt{x}, \qquad g(x) = \sqrt[3]{x} \]
The difference is the parity of the index, exactly as in Lesson 6.1: an even root of a negative number does not exist, and an odd root of one does.
Figure (svg): The square root and cube root parent functions graphed side by side with their domains and ranges
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.5 Graph Square Root and Cube Root Functions §6.5, pp. 446-446 — Parent Functions for Square Root and Cube Root Functions
Picture it
Each passes through the origin and through the point where x and y are both 1.
Figure (svg): The square root and cube root parent functions graphed side by side with their domains and ranges
The cube root also passes through the point where both coordinates are negative 1, which the square root cannot, because negative 1 is not in its domain.
Worked example
The two graphs, read off carefully.
\[ \text{State the domain, range and shape of } y = \sqrt{x} \text{ and } y = \sqrt[3]{x}. \]
Square root: find the domain
Why: A square root needs a non-negative radicand, so x must be at least zero.
\[ \text{domain } x \ge 0 \]
Square root: find the range
Why: The principal square root is never negative, so the outputs are at least zero.
\[ \text{range } y \ge 0 \]
Cube root: find the domain
Why: An odd index accepts every real number, including negatives.
Cube root: find the range
Why: Every real number is the cube root of something, so every real value is attained.
Figure (svg): The solution to Worked example describe both parents shown as a ladder of expressions, one row per algebraic move
\[ \sqrt{x}: \; x \ge 0, y \ge 0; \qquad \sqrt[3]{x}: \; \text{all reals} \]
Verify: check against the inverse relationship
Why: The square root is the inverse of x squared restricted to non-negatives, whose domain and range are both non-negative — and an inverse swaps those two sets, so both stay non-negative. The cube root inverts x cubed, whose domain and range are both all reals. The graphs agree with the algebra of Lesson 6.4.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.5 Graph Square Root and Cube Root Functions §6.5, pp. 446-446
Sorting
Check whether a real root exists.
Sort into buckets
Sort each point.
The two shared points are worth memorising: they anchor both parent sketches, and every transformation can be tracked by following them.
Worked example
Three test points, and what they show.
\[ \text{Which of } (0,0), (1,1), (-1,-1), (4,2), (-8,-2) \text{ lie on } y=\sqrt{x} \text{ and on } y=\sqrt[3]{x}? \]
Test the origin
Why: Both roots of zero are zero, so the origin is on both.
Test the point where both are 1
Why: Both roots of 1 are 1.
Test the point with both coordinates negative 1
Why: The cube root of negative 1 is negative 1; the square root does not exist.
Test the remaining two
Why: The square root of 4 is 2; the cube root of negative 8 is negative 2.
Figure (svg): The solution to Worked example which points are on each graph shown as a ladder of expressions, one row per algebraic move
\[ \text{shared: } (0,0), (1,1) \]
Verify: notice what the shared points mean
Why: Both graphs pass through the origin and through the point where both coordinates are 1, because zero and one are their own roots at every index. Those two points are the natural anchors for sketching either curve, and every transformation in this lesson can be tracked by watching where they move to.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.5 Graph Square Root and Cube Root Functions §6.5, pp. 446-446
Trap
\[ y = \sqrt{x} \]
Sketch it symmetrically about the vertical axis
Why: The curve is drawn on both sides of the origin, like a parabola.
\[ \text{a left branch through } (-4, 2) \quad \text{(wrong)} \]
The square root of negative 4 does not exist among the real numbers, so there is no point of the graph anywhere to the left of the origin.
\[ \text{domain of } \sqrt{x} \text{ is } x \ge 0 \]
Draw only the right half, starting at the origin
Why: The graph is half of a sideways parabola, not a whole one.
\[ \sqrt[3]{x} \text{ does continue left, because } 3 \text{ is odd} \]
The two parents differ in exactly this way, and confusing them is the most common error in the lesson. Check the parity of the index before drawing anything.
Matching
The parity of the index decides both.
Match the pairs
Why: The two even indices behave alike and the two odd ones behave alike, whatever the actual index is. Only the parity matters, which is why this lesson needs just two parent functions rather than one per index.
Fill the middle
The square root parent.
Fill in the blanks
y = \sqrt0: \; \text___ \ge 0, \text___ x \ge ___
Why: The radicand here is x itself, so requiring it to be non-negative gives x at least zero directly. When the radicand is more complicated, as in the fourth idea, the same condition produces a different inequality to solve.
Prediction
Commit before reasoning.
Predict first
Why is the cube root of a negative number defined when the square root is not?
Correct: Because an odd number of negative factors leaves a negative product.
\[ (-2)^3 = -8 \quad \text{but} \quad b^2 \ge 0 \text{ for all real } b \]
Why: Negative 2 cubed is negative 8, since three negative factors leave one unpaired. With an even index the factors pair off completely and the result is never negative, so no real number squares to a negative. This is the same parity argument from Lesson 6.1, appearing here as the difference between a graph with an endpoint and one without.
Section
Section 2
Concept
In y equals a times a root of x, the size of a stretches the graph vertically when it exceeds 1 and shrinks it when it is between 0 and 1. The sign of a reflects the graph across the horizontal axis when it is negative.
\[ y = \tfrac{1}{2}\sqrt{x} \text{ shrinks}; \qquad y = -3\sqrt[3]{x} \text{ stretches and flips} \]
The coefficient never changes the domain, because it acts after the root has been taken. For a square root it does change the range's direction: a negative a turns y at least zero into y at most zero.
Figure (svg): A square root shrunk vertically and a cube root stretched and reflected
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.5 Graph Square Root and Cube Root Functions §6.5, pp. 446-447 — Graph a square root function
Picture it
Examples 1 and 2, each with its parent dashed behind it.
Figure (svg): A square root shrunk vertically and a cube root stretched and reflected
In the first, every height is halved; in the second, every height is tripled and its sign reversed. The horizontal positions are untouched in both.
Worked example
Examples 1 and 2, with domains and ranges.
\[ \text{Graph } y = \tfrac{1}{2}\sqrt{x} \text{ and } y = -3\sqrt[3]{x}, \text{ and compare each with its parent.} \]
First: tabulate a few values
Why: At x equal to 0, 1, 4 the outputs are 0, 0.5 and 1.
First: describe the change
Why: Every output is half the parent's, so it is a vertical shrink by a factor of one half.
First: state the domain and range
Why: The radicand is still x, so the domain is unchanged, and halving non-negative values keeps them non-negative.
\[ x \ge 0, y \ge 0 \]
Second: tabulate
Why: At x equal to negative 1, 0, 1 the outputs are 3, 0 and negative 3.
Second: describe and state
Why: A vertical stretch by 3 followed by a reflection in the horizontal axis; the domain and range are all reals.
Figure (svg): The solution to Worked example a shrink and a reflected stretch shown as a ladder of expressions, one row per algebraic move
\[ y = \tfrac{1}{2}\sqrt{x}; \qquad y = -3\sqrt[3]{x} \]
Verify: check one point on each
Why: For the first, at x equal to 4 the parent gives 2 and the shrunk graph gives 1 — exactly half. For the second, at x equal to 1 the parent gives 1 and the transformed graph gives negative 3, which is 3 times 1 with the sign reversed. Both changes are vertical only, and the x values were never touched.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.5 Graph Square Root and Cube Root Functions §6.5, pp. 446-447
Sorting
Read the size and the sign of a.
Sort into buckets
Sort each coefficient by its effect.
Two independent readings from one number, exactly as for the leading coefficient of a parabola in Lesson 4.1.
Worked example
Guided Practice 1 to 4.
\[ \text{Graph } y = -3\sqrt{x}, \; f(x) = \tfrac{1}{4}\sqrt{x}, \; y = -\tfrac{1}{2}\sqrt[3]{x}, \; g(x) = 4\sqrt[3]{x}. \]
First: a negative stretch of a square root
Why: Stretched by 3 and flipped, so the outputs are at most zero.
\[ x \ge 0, y \le 0 \]
Second: a shrink of a square root
Why: Shrunk by one quarter, with the signs unchanged.
\[ x \ge 0, y \ge 0 \]
Third: a negative shrink of a cube root
Why: Shrunk by one half and flipped.
Fourth: a stretch of a cube root
Why: Stretched by 4, with no flip.
Figure (svg): The solution to Worked example four more shown as a ladder of expressions, one row per algebraic move
\[ y \le 0; \; y \ge 0; \; \text{all reals}; \; \text{all reals} \]
Verify: notice which ranges the sign changed
Why: For the two square root functions, the sign of a flipped the range from y at least zero to y at most zero. For the two cube root functions the range was already all real numbers, so flipping changed nothing about it. The coefficient's effect on the range depends on which parent it is applied to.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.5 Graph Square Root and Cube Root Functions §6.5, pp. 447-447
Error analysis
A student states the domain of a transformed square root function.
Annotate
On: \( y = -3\sqrt{x}: \text{ the negative sign means } x \le 0 \)
The domain comes from the radicand and the range from everything outside it. Keeping the two questions separate makes both easy.
Fill the middle
Example 2, at x equal to 1.
Fill in the blanks
y = -3\sqrt[3]-3: \; \text___ x = 1, \; y = -3(1) = ___
Why: The cube root of 1 is 1, and multiplying by negative 3 gives negative 3. The parent passed through the point where both coordinates are 1; the transformed graph passes through the point 1 across and 3 down, which is the stretch and the flip together.
Comparison
Fill the blanks. The coefficient touches only one of them.
Comparison matrix
| Question | Square root | Cube root |
|---|---|---|
| Does a change the domain? | no | no |
| Range when a > 0 | y >= 0 | all reals |
| Range when a < 0 | y <= 0 | all reals |
| Why the difference | the parent's range is one-sided | the parent's range is already everything |
Flipping a one-sided range moves it to the other side; flipping a range that is already all real numbers changes nothing.
Prediction
Commit before reasoning.
Predict first
Why does the coefficient never affect which inputs are allowed?
Correct: Because it acts after the root is taken.
\[ y = a\sqrt{x}: \; \text{the domain comes from } \sqrt{x}, \text{ never from } a \]
Why: The order of operations puts the root first: an input goes into the radicand, the root is taken, and only then is the result multiplied by a. So whatever a is, it cannot rescue an input the radicand rejected or reject one the radicand accepted. Anything inside the radical can change the domain, and nothing outside it can.
Section
Section 3
Concept
To graph a times the square root of x minus h, plus k, sketch a times the square root of x first, then shift it h units horizontally and k units vertically. The starting point moves from the origin to the point with coordinates h and k.
\[ y = a\sqrt{x-h}+k \]
The domain becomes x at least h, because the radicand x minus h must be non-negative. The range becomes y at least k, or y at most k when a is negative.
Figure (svg): A square root function translated right and up, with its starting point marked
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.5 Graph Square Root and Cube Root Functions §6.5, pp. 448-448 — Graph a translated square root function
Picture it
Example 4: negative 2 times the square root of x minus 3, plus 2.
Figure (svg): A square root function translated right and up, with its starting point marked
The endpoint moved from the origin to the point 3 across and 2 up, and the whole curve came with it. The domain and range shifted by exactly the same amounts.
Worked example
Example 4, in the book's two steps.
\[ \text{Graph } y = -2\sqrt{x-3}+2 \text{ and state the domain and range.} \]
Sketch the untranslated version
Why: Negative 2 times the square root of x starts at the origin and passes through the point 1 across and 2 down.
\[ y = -2 \sqrt{x} \]
Read h and k
Why: The form subtracts h, so x minus 3 gives h equal to 3; the constant outside is k, which is 2.
\[ h = 3, k = 2 \]
Shift the graph
Why: Right 3 and up 2, so the endpoint moves to the point 3 across and 2 up.
\[ \text{starts at } (3, 2) \]
State the domain and range
Why: The radicand needs x at least 3; the negative coefficient makes the outputs at most 2.
\[ x \ge 3, y \le 2 \]
Figure (svg): The solution to Worked example translate and describe shown as a ladder of expressions, one row per algebraic move
\[ x \ge 3, \quad y \le 2 \]
Verify: check a second point
Why: The untranslated graph passed through the point 1 across and 2 down; shifting right 3 and up 2 sends it to the point 4 across and 0 up. Substituting x equal to 4 gives negative 2 times the square root of 1, plus 2, which is 0. The point is on the graph, so the translation was applied correctly.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.5 Graph Square Root and Cube Root Functions §6.5, pp. 448-448
Matching
The graph starts at the point with coordinates h and k.
Match the pairs
Why: Two of these share h equal to 3 and differ only in k, and two share k equal to 2 and differ only in h. Reading the two numbers independently is what makes the starting point immediate.
Worked example
Guided Practice 6 to 8.
\[ \text{Graph } y = -4\sqrt{x}+2, \; y = 2\sqrt{x+1}, \; f(x) = \tfrac{1}{2}\sqrt{x-3}-1. \]
First: no horizontal shift
Why: H is 0 and k is 2, with a negative coefficient.
\[ x \ge 0, y \le 2 \]
Second: a left shift
Why: X plus 1 means h is negative 1, and k is 0, with a positive coefficient.
\[ x \ge - 1, y \ge 0 \]
Third: right and down
Why: H is 3 and k is negative 1, with a positive coefficient.
\[ x \ge 3, y \ge - 1 \]
Note the pattern
Why: The domain always begins at h and the range always begins at k.
Figure (svg): The solution to Worked example three more square root translations shown as a ladder of expressions, one row per algebraic move
\[ (0, 2); \quad (-1, 0); \quad (3, -1) \]
Verify: check the second's endpoint
Why: At x equal to negative 1 the radicand is zero, so y is zero and the graph starts at the point negative 1 across and 0 up — which is the point with coordinates h and k. Every one of these starts at that point, which is the fastest single check on a translated radical graph.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.5 Graph Square Root and Cube Root Functions §6.5, pp. 449-449
Trap
\[ y = 2\sqrt{x+1} \]
Take h as positive 1 because the radicand shows a plus
Why: The sign inside is copied straight into the shift.
\[ \text{shift right } 1, \text{ domain } x \ge 1 \quad \text{(wrong)} \]
At x equal to 0 the radicand is 1 and the function has the value 2, so 0 is in the domain — which x at least 1 would exclude.
\[ x+1 = x-(-1) \;\Longrightarrow\; h = -1 \]
Ask what makes the radicand zero
Why: That value is h, and the graph starts there.
\[ \text{shift left } 1, \text{ domain } x \ge -1 \]
The same sign trap as the vertex form of Lesson 4.2 and the absolute value form of Lesson 2.7. Asking what makes the bracket vanish avoids it in every one of those settings.
Fill the middle
Example 4.
Fill in the blanks
y = -2\sqrt3+2: \; x - 3 \ge 0 \;\Longrightarrow\; x \ge ___
Why: Requiring the radicand to be non-negative gives x at least 3, which is the same value as h. The domain of a translated square root function always begins exactly at h, because that is where the radicand is zero.
Sorting
The sign of a decides.
Sort into buckets
Sort each function by its range.
The value k is always the boundary; the sign of a says which side of it the graph occupies.
Prediction
Commit before reasoning.
Predict first
Why does the graph of a square root function begin exactly at x equal to h?
Correct: Because the radicand is zero there and negative to the left.
\[ x < h \;\Longrightarrow\; x-h < 0 \;\Longrightarrow\; \sqrt{x-h} \text{ not real} \]
Why: For any x smaller than h the expression x minus h is negative, and a square root of a negative number is not real. At x equal to h the radicand is zero, which is allowed, so that is the leftmost point of the graph. The translation description and the radicand description agree, and the second is the one that proves it.
Section
Section 4
Concept
A cube root function is translated in exactly the same way: sketch a times the cube root of x, then shift h horizontally and k vertically. Because the parent's domain and range are already all real numbers, translating changes neither.
\[ y = a\sqrt[3]{x-h}+k \]
There is no endpoint to track, so instead follow three points: where the parent passes through the origin and through the two points one unit either side of it.
Figure (svg): A cube root function translated left and down, with three matching points marked on each graph
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.5 Graph Square Root and Cube Root Functions §6.5, pp. 448-448 — Graph a translated cube root function
Picture it
Example 5: 3 times the cube root of x plus 4, minus 1.
Figure (svg): A cube root function translated left and down, with three matching points marked on each graph
Three marked points on the parent moved to three marked points on the image, each shifted 4 left and 1 down. Nothing about the domain or the range changed.
Worked example
Example 5, in the book's two steps.
\[ \text{Graph } y = 3\sqrt[3]{x+4}-1 \text{ and state the domain and range.} \]
Sketch the untranslated version
Why: Three times the cube root of x passes through the origin and through the points one unit either side, at heights 3 and negative 3.
Read h and k
Why: X plus 4 means h is negative 4, and the constant outside is negative 1.
\[ h = -4, k = -1 \]
Shift the graph
Why: Left 4 and down 1.
\[ \text{left } 4,\text{ down } 1 \]
Track the anchor points and state domain and range
Why: The three points move to 4 left and 1 down of where they were; the domain and range remain all real numbers.
Figure (svg): The solution to Worked example translate a cube root function shown as a ladder of expressions, one row per algebraic move
\[ (-4,-1), \; (-3,2), \; (-5,-4) \]
Verify: substitute one point
Why: At x equal to negative 3 the radicand is 1, whose cube root is 1, so y is 3 minus 1, which is 2 — matching the marked point. Checking one shifted anchor confirms both the direction and the size of the translation at once.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.5 Graph Square Root and Cube Root Functions §6.5, pp. 448-448
Fill the middle
Example 5.
Fill in the blanks
y = 3\sqrt[3]-4-1: \; x+4 = x-(___) \;\Longrightarrow\; h = -4
Why: The radicand vanishes at x equal to negative 4, so h is negative 4 and the shift is 4 units to the left. Asking what makes the radicand zero gives h directly, without any sign bookkeeping.
Worked example
Guided Practice 9 to 11.
\[ \text{Graph } y = -\sqrt[3]{x-4}, \; y = \sqrt[3]{x}-5, \; g(x) = -\sqrt[3]{x+2}-3. \]
First: right 4, and flipped
Why: H is 4 and k is 0, with a negative coefficient.
\[ \text{through } (4, 0) \]
Second: down 5
Why: H is 0 and k is negative 5, with a positive coefficient.
\[ \text{through } (0, -5) \]
Third: left 2 and down 3, flipped
Why: H is negative 2 and k is negative 3.
\[ \text{through } (-2, -3) \]
State every domain and range
Why: All three are cube root functions, so all six answers are the same.
Figure (svg): The solution to Worked example three more cube root translations shown as a ladder of expressions, one row per algebraic move
\[ \text{all reals, in every case} \]
Verify: check that the flips changed nothing about the range
Why: Two of the three have a negative coefficient, which reflects the graph across the horizontal axis — but the range was already all real numbers and stays that way. For a cube root function, a and h and k move and reshape the curve without ever restricting it.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.5 Graph Square Root and Cube Root Functions §6.5, pp. 449-449
Error analysis
A student states the domain of a translated cube root function.
Annotate
On: \( y = 3\sqrt[3]{x+4}-1: \text{ the radicand needs } x+4 \ge 0, \text{ so } x \ge -4 \)
Check the parity of the index before writing any domain condition. Only an even index restricts the radicand.
Sorting
Even index restricts; odd index does not.
Sort into buckets
Sort each function by whether its domain is restricted.
The value of h shifts the graph in both cases; only for an even index does it also become the edge of the domain.
Comparison
Fill the blanks. Same shifts, different consequences.
Comparison matrix
| Question | a sqrt(x - h) + k | a cbrt(x - h) + k |
|---|---|---|
| Horizontal shift | h units | h units |
| Vertical shift | k units | k units |
| Domain | x >= h | all real numbers |
| Range | y >= k or y <= k | all real numbers |
The transformations behave identically; only the parent's domain and range make the consequences differ.
Prediction
Commit before reasoning.
Predict first
A cube root graph has no endpoint. What is the best set of points to follow through a translation?
Correct: The origin and the two points one unit either side of it.
\[ (0,0), (1,a), (-1,-a) \;\to\; (h,k), (h+1, k+a), (h-1, k-a) \]
Why: On the parent those three points are easy to compute and well spread, and after a stretch by a they sit at heights 0, a and negative a. Following them through the shift pins the curve's position and its steepness at once. The origin plays the role the endpoint plays for a square root: it is the centre of the shape rather than its edge.
Section
Section 5
Concept
A square root model rises quickly at first and then flattens, because the outputs grow with the square root of the input. Reading a value from its graph, or solving algebraically, answers the question the model was built for.
\[ T = 1.11\sqrt{l} \]
The flattening is the practical content of a square root model: quadrupling the input only doubles the output, so large inputs buy very little extra.
Figure (svg): A pendulum's period plotted against its length, with the length giving a three-second period marked
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.5 Graph Square Root and Cube Root Functions §6.5, pp. 447-447 — Solve a multi-step problem
Picture it
Example 3: the period in seconds against the length in feet.
Figure (svg): A pendulum's period plotted against its length, with the length giving a three-second period marked
A three-second period needs about 7.3 feet. Doubling that length to 14.6 feet would give a period of only about 4.2 seconds, not 6.
Worked example
Example 3 and Guided Practice 5.
\[ \text{With } T = 1.11\sqrt{l}, \text{ find the length giving a period of } 3 \text{ seconds, and of } 1 \text{ second.} \]
Graph the model over sensible lengths
Why: Only non-negative lengths make sense, so the graph starts at the origin.
\[ l \ge 0 \]
Trace to where the period is 3
Why: The graph reaches a height of 3 at about 7.3.
\[ \text{about } 7.3\text{ feet} \]
Solve the second algebraically
Why: One equals 1.11 times the square root of l, so the square root of l is about 0.901.
\[ \sqrt{l}\text{ about } 0.901 \]
Square both sides
Why: Squaring 0.901 gives about 0.81.
\[ \text{about } 0.81\text{ feet} \]
Figure (svg): The solution to Worked example read a length from the graph shown as a ladder of expressions, one row per algebraic move
\[ 7.3 \text{ ft}; \qquad 0.81 \text{ ft} \]
Verify: substitute back
Why: At 7.3 feet the square root is about 2.702, and 1.11 times that is about 3.00 seconds. At 0.81 feet the square root is 0.9, and 1.11 times that is about 1.00 second. Both check, and note how much less length the shorter period needed: a third of the period took only about a ninth of the length.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.5 Graph Square Root and Cube Root Functions §6.5, pp. 447-447
Fill the middle
Guided Practice 5.
Fill in the blanks
1 = 1.11\sqrt0.81 \;\Longrightarrow\; \sqrt___ = 0.901 \;\Longrightarrow\; l \approx ___
Why: Squaring 0.901 gives about 0.81 feet, which is roughly 9.7 inches. Solving a square root equation by squaring both sides is the subject of Lesson 6.6, where the danger of doing so is examined properly.
Worked example
The scaling behind the model, checked at three lengths.
\[ \text{Find } T \text{ at } l = 1, 4 \text{ and } 16, \text{ and describe the pattern.} \]
Evaluate at 1 foot
Why: The square root of 1 is 1, so the period is 1.11 seconds.
\[ 1.11 s \]
Evaluate at 4 feet
Why: The square root of 4 is 2, so the period is 2.22 seconds.
\[ 2.22 s \]
Evaluate at 16 feet
Why: The square root of 16 is 4, so the period is 4.44 seconds.
\[ 4.44 s \]
Describe the pattern
Why: Each quadrupling of the length doubles the period.
Figure (svg): The solution to Worked example how the period responds shown as a ladder of expressions, one row per algebraic move
\[ 1.11, \; 2.22, \; 4.44 \text{ seconds} \]
Verify: check the scaling algebraically
Why: Replacing l by 4l gives 1.11 times the square root of 4l, which is 1.11 times 2 times the square root of l — exactly twice the original period. The factor 2 is the square root of 4, which is the same square-root scaling as the coral cod in Lesson 6.1 and the balloon radius in this chapter's transfer problems.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.5 Graph Square Root and Cube Root Functions §6.5, pp. 447-447
Trap
\[ T = 1.11\sqrt{l}: \; \text{at } l = 7.3, \; T = 3 \]
Double the length to double the period
Why: The relationship is treated as proportional.
\[ l = 14.6 \;\Longrightarrow\; T = 6 \quad \text{(wrong)} \]
At 14.6 feet the square root is about 3.82, so the period is about 4.24 seconds — not 6.
\[ T(4l) = 1.11\sqrt{4l} = 2\cdot 1.11\sqrt{l} = 2T(l) \]
Scale by the square root of the factor, not by the factor
Why: Doubling the length multiplies the period by the square root of 2, about 1.41.
\[ l = 29.2 \;\Longrightarrow\; T \approx 6 \]
To double a period you must quadruple the length. Every square root model behaves this way, and the flattening of the graph is the same fact drawn.
Prediction
Commit before reasoning.
Predict first
A pendulum has a 2 second period. What length gives a 4 second period?
Correct: Four times as long.
\[ T \propto \sqrt{l} \;\Longleftrightarrow\; l \propto T^2 \]
Why: The period is proportional to the square root of the length, so the length is proportional to the square of the period. Doubling the period squares to a factor of 4 on the length. This is Lesson 6.4's inverse relationship in action: inverting T equals 1.11 root l gives l equal to T squared over 1.2321, and squares grow much faster than square roots.
Sorting
Does the quantity grow faster or slower than its input?
Sort into buckets
Sort each relationship by its shape.
Each pair here consists of a function and its inverse, which is why one of each pair is in each bucket.
Ranking
Answering a question with a radical model.
Put in order
Why: Step two matters here because a square root model already restricts its own domain, and the situation usually restricts it further — a length cannot be negative, and the model would not be trusted for a mile-long pendulum either. Step five's units are part of the answer, not decoration.
Comparison
Fill the blanks. Every family in this course uses a, h and k the same way.
Comparison matrix
| Family | Form | What h and k do |
|---|---|---|
| Absolute value, 2.7 | a|x - h| + k | move the vertex to (h, k) |
| Quadratic, 4.2 | a(x - h)^2 + k | move the vertex to (h, k) |
| Square root, 6.5 | a sqrt(x - h) + k | move the endpoint to (h, k) |
| Cube root, 6.5 | a cbrt(x - h) + k | move the centre point to (h, k) |
Learning one family's transformations teaches all of them. Only the parent shape and its domain change from row to row.
Pattern
One routine for graphing any radical function.
For a model, restrict the domain to the values the situation allows, and remember that a square root's output grows by the square root of whatever factor the input grows by.
OpenStax Algebra and Trigonometry 2e, §5.7 Inverses and Radical Functions §5.7
Check
The parent functions.
Check your understanding
What are the domain and range of y = cube root of x?
Answer: A
Why: An odd index accepts every real radicand and produces every real value.
Check
A translation. Watch the sign of h.
Check your understanding
What are the domain and range of y = -2 sqrt(x - 3) + 2?
Answer: A
Why: The radicand needs x >= 3, and the negative coefficient sends the outputs downward from k = 2.
Check
A model. Square root scaling.
Check your understanding
With T = 1.11 sqrt(l), a 7.3 foot pendulum has a 3 second period. What length gives a 6 second period?
Answer: A
Why: Doubling the period requires quadrupling the length, and 4 times 7.3 is 29.2.
Real world
The distance in miles to the visible horizon from a height of h feet is about 1.22 times the square root of h.
Discussion prompt
Graph the model, find the horizon distance from a 100-foot lighthouse, and find the height needed to see 20 miles. Then say how much extra height doubles the horizon distance.
Hint: The model is a square root function with a equal to 1.22, h equal to 0 and k equal to 0.
Answer:
\[ d = 1.22\sqrt{h}: \; d(100) = 1.22(10) = 12.2 \text{ miles} \]
\[ 20 = 1.22\sqrt{h} \;\Longrightarrow\; \sqrt{h} \approx 16.39 \;\Longrightarrow\; h \approx 269 \text{ feet} \]
A 100-foot lighthouse sees about 12.2 miles, and seeing 20 miles needs about 269 feet. Doubling the horizon distance requires four times the height.
Two things are worth noticing. The graph starts at the origin and flattens, which is why the second lighthouse needs more than twice the height for less than twice the view — the same square-root scaling as the pendulum. And the domain restriction is supplied by the situation as well as by the mathematics: a negative height is meaningless, and the square root would not accept it anyway.
Commit first
Answer, then rate your confidence honestly.
Predict first
Does the graph of y equals the cube root of x plus 4 have a restricted domain?
Correct: No — an odd index accepts every real radicand.
\[ \sqrt[3]{-8} = -2 \;\Longrightarrow\; \text{no restriction on the radicand} \]
Why: The cube root of negative 8 is negative 2, so a negative radicand is perfectly acceptable. The value negative 4 is still meaningful: it is h, the point the graph is centred on after the shift. But being the centre is not the same as being an endpoint, and only an even index produces an endpoint. This is the single most useful distinction in the lesson, and it comes straight from Lesson 6.1's parity rule.
Explain it
They can graph parabolas from vertex form and have just met a square root function.
Discussion prompt
In four sentences or fewer, explain how to graph a times the square root of x minus h, plus k, by comparing it with what they already know.
Hint: The letters do the same jobs.
Answer:
The letters mean exactly what they meant in vertex form: a stretches the graph and flips it if it is negative, h shifts it horizontally and k vertically. The only difference is the shape being moved — half a sideways parabola instead of a whole upright one.
Sketch a times the square root of x first, which starts at the origin, then shift it so that its starting point lands at the point with coordinates h and k. The domain begins at h because the radicand cannot be negative, and the range begins at k and runs whichever way the sign of a points.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the parents, check the parity of the index before drawing anything. For h, ask what makes the radicand zero. For the range, note that k is always the boundary and the sign of a says which side. For domains, only an even index restricts anything. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Take y equals negative 2 times the square root of x minus 3, plus 2, and y equals 3 times the cube root of x plus 4, minus 1, and put them side by side. For each: write down a, h and k, sketch the untranslated version first with its anchor points marked, then draw the translated graph with those points shifted, and state the domain and range with a sentence saying where each came from. Below both, write the one-sentence rule explaining why one has an endpoint and the other does not. At the bottom, sketch the pendulum model and mark the lengths giving periods of 1, 2 and 4 seconds, noting the factor between consecutive lengths. In a margin, list the four function families that use a, h and k in the same way.
If your two length factors are not 4 each time, recheck: doubling a square root's output always needs four times the input.
Recap
Five things, and the parity of the index explains most of them.
| If you see | Then |
|---|---|
| An even index | Endpoint, restricted domain |
| An odd index | No endpoint, all real numbers |
| A coefficient bigger than 1 | Vertical stretch |
| A negative coefficient | Reflection across the horizontal axis |
| x minus h in the radicand | Shift right h, domain starts at h |
| A constant k outside | Shift vertically, range starts at k |
| A square root model | Growth slows: quadruple the input to double the output |
Lesson 6.6 closes the chapter by solving equations that contain radicals, where squaring both sides is the natural move and also the source of a genuinely new hazard.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.5 Graph Square Root and Cube Root Functions §6.5, pp. 446-451 — everything on these slides traces back here
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