Building and transforming functions

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1. Composition of functions

Warm-up

Three quick questions from earlier lessons.

Discover

gg doubles a number: g(x)=2xg(x) = 2x. ff adds 1: f(x)=x+1f(x) = x + 1. Start with 3: double it, then add 1. You have computed f(g(3))f(g(3)). What is it?

  1. 7
  2. 8
  3. 24

Key idea

Composing means using the output of one function as the input of another.

f(g(x))f(g(x)): first apply gg to xx, then apply ff to the result. Work from the inside out. In general f(g(x))≠g(f(x))f(g(x)) \ne g(f(x)).

Worked example

f(x)=x2f(x) = x^2 and g(x)=x−3g(x) = x - 3. Find f(g(5))f(g(5)), then f(g(x))f(g(x)).

g(5)=5−3=2g(5) = 5 - 3 = 2
The inside first.
f(g(5))=f(2)=4f(g(5)) = f(2) = 4
Then f of that result.
f(g(x))=(x−3)2f(g(x)) = (x - 3)^2
With x: put g(x) = x − 3 in place of x in f.

Your turn, guided

Your turn: f(x)=2x+1f(x) = 2x + 1 and g(x)=x2g(x) = x^2. Find f(g(3))f(g(3)).

g(3)=9g(3) = 9
The inside first.
f(g(3))=f(9)=19f(g(3)) = f(9) = 19
Then 2 · 9 + 1.

Practice ★

  1. f(x)=2x+2f(x) = 2x + 2 and g(x)=x2−2g(x) = x^2 - 2. Find f(g(3))f(g(3)).
  2. f(x)=3x+3f(x) = 3x + 3 and g(x)=x2−4g(x) = x^2 - 4. Find f(g(3))f(g(3)).
  3. f(x)=4x+3f(x) = 4x + 3 and g(x)=x2−1g(x) = x^2 - 1. Find f(g(2))f(g(2)).
  4. f(x)=3x+1f(x) = 3x + 1 and g(x)=x2−5g(x) = x^2 - 5. Find f(g(3))f(g(3)).

Watch out!

Practice ★★

  1. f(x)=6x+6f(x) = 6x + 6 and g(x)=x+5g(x) = x + 5. Write f(g(x))f(g(x)).
  2. f(x)=5x+7f(x) = 5x + 7 and g(x)=x+2g(x) = x + 2. Write f(g(x))f(g(x)).
  3. f(x)=6x+8f(x) = 6x + 8 and g(x)=x+3g(x) = x + 3. Write f(g(x))f(g(x)).
  1. g(1)=5g(1) = 5, g(5)=7g(5) = 7, f(1)=7f(1) = 7 and f(5)=1f(5) = 1. Find f(g(1))f(g(1)).
  2. g(2)=6g(2) = 6, g(6)=7g(6) = 7, f(2)=7f(2) = 7 and f(6)=2f(6) = 2. Find f(g(2))f(g(2)).

Exit ticket

Three questions to check you have got it.

  1. f(x)=2x+2f(x) = 2x + 2 and g(x)=x2−2g(x) = x^2 - 2. Find f(g(3))f(g(3)).
  2. f(x)=3x+3f(x) = 3x + 3 and g(x)=x2−4g(x) = x^2 - 4. Find f(g(3))f(g(3)).
  3. f(x)=4x+3f(x) = 4x + 3 and g(x)=x2−1g(x) = x^2 - 1. Find f(g(2))f(g(2)).

Summary card

Composition

  • f(g(x))f(g(x)): apply gg first, then ff.
  • Work from the inside out.
  • The order matters: f(g(x))f(g(x)) and g(f(x))g(f(x)) usually differ.

Watch out: f(g(x))f(g(x)) is not f(x)⋅g(x)f(x) \cdot g(x).

2. Transformations: shifts and reflections

Warm-up

Three quick questions from earlier lessons.

Discover

Move the sliders and watch the graph of y=(x−h)2+ky = (x - h)^2 + k.

The graph of y = (x − h)² + k, a parabola whose vertex is at (h, k) and moves with the sliders.

When hh goes from 0 to 2, which way does the graph move?

  1. To the right
  2. To the left

Key idea

Changing the rule moves or flips the graph without changing its shape.

y=f(x−h)+ky = f(x - h) + k moves the graph of ff by hh units to the right and kk units up. y=−f(x)y = -f(x) reflects it across the xx-axis; y=f(−x)y = f(-x) reflects it across the yy-axis.

A change inside f(… )f(\dots) acts on xx (horizontally); a change outside acts on yy (vertically).

Worked example

Describe y=(x+1)2−3y = (x + 1)^2 - 3 compared with y=x2y = x^2.

1 unit to the left
x + 1 = x − (−1): h = −1.
3 units down
− 3 outside: k = −3.
Vertex: (−1,−3)(-1, -3)
The vertex (0, 0) moves with the graph.

Your turn, guided

Your turn: y=(x−4)2+2y = (x - 4)^2 + 2.

4 units to the right
x − 4: h = 4.
2 units up
+ 2 outside.
The vertex is at x=4x = 4
The vertex moves from x = 0 to x = …

Practice ★

  1. The graph of y=x2y = x^2 is moved 1 units to the right and 6 units down. Write the equation of the new graph.
  2. The graph of y=x2y = x^2 is moved 3 units to the right and 1 units down. Write the equation of the new graph.
  3. The graph of y=x2y = x^2 is moved 4 units to the right and 6 units down. Write the equation of the new graph.
  4. The graph of y=x2y = x^2 is moved 7 units to the right and 1 units down. Write the equation of the new graph.

Watch out!

Practice ★★

  1. Draw the graph of y=(x−1)2−3y = (x - 1)^2 - 3: place the vertex and one more point.
  2. Draw the graph of y=(x−1)2−1y = (x - 1)^2 - 1: place the vertex and one more point.
  1. Match each equation with what it does to the graph of y=f(x)y = f(x).
    1. y=f(x)+3y = f(x) + 3
    2. y=f(x−3)y = f(x - 3)
    3. y=−f(x)y = -f(x)
    4. y=f(−x)y = f(-x)
    1. moves it up 3 units
    2. moves it 3 units to the right
    3. reflects it across the x-axis
    4. reflects it across the y-axis

Exit ticket

Three questions to check you have got it.

  1. The graph of y=x2y = x^2 is moved 1 units to the right and 6 units down. Write the equation of the new graph.
  2. The graph of y=x2y = x^2 is moved 3 units to the right and 1 units down. Write the equation of the new graph.
  3. The graph of y=x2y = x^2 is moved 4 units to the right and 6 units down. Write the equation of the new graph.

Summary card

Shifts and reflections

  • f(x−h)f(x - h): hh units right. f(x)+kf(x) + k: kk units up.
  • −f(x)-f(x): flip across the xx-axis. f(−x)f(-x): flip across the yy-axis.
  • Inside f(… )f(\dots) acts on xx; outside acts on yy.

Watch out: f(x+3)f(x + 3) moves the graph left.

3. Stretches and compressions; even and odd functions

Warm-up

Three quick questions from earlier lessons.

Discover

Move the slider and watch the graph of y=(bx)2y = (bx)^2.

The graph of y = (bx)², a parabola that gets narrower as b grows.

When bb goes from 1 to 2, the graph becomes…

  1. narrower
  2. wider

Key idea

Multiplying stretches or squeezes a graph. Symmetry has a name.

For a>1a > 1: y=af(x)y = a f(x) stretches the graph vertically by aa. For b>1b > 1: y=f(bx)y = f(bx) compresses it horizontally by bb. ff is even if f(−x)=f(x)f(-x) = f(x) (symmetric about the yy-axis), odd if f(−x)=−f(x)f(-x) = -f(x) (symmetric about the origin).

Worked example

Is f(x)=x3−2xf(x) = x^3 - 2x even, odd, or neither?

f(−x)=(−x)3−2(−x)f(-x) = (-x)^3 - 2(-x)
Replace x by (−x).
=−x3+2x= -x^3 + 2x
(−x)³ = −x³ and −2(−x) = 2x.
=−(x3−2x)=−f(x)= -(x^3 - 2x) = -f(x): odd
Factor out −1: this is −f(x).

Your turn, guided

Your turn: f(x)=x4+3f(x) = x^4 + 3. Check one value.

f(−1)=(−1)4+3f(-1) = (-1)^4 + 3
Replace x by −1.
f(−1)=4f(-1) = 4
(−1)⁴ = 1. It equals f(1): a sign that f is even.

Practice ★

  1. How does the graph of y=f(4x)y = f(4x) compare with the graph of y=f(x)y = f(x)?
    1. It is squeezed toward the y-axis: horizontal compression by a factor of 4
    2. It is pulled away from the y-axis: horizontal stretch by a factor of 4
    3. It is stretched vertically by a factor of 4
  2. How does the graph of y=f(5x)y = f(5x) compare with the graph of y=f(x)y = f(x)?
    1. It is squeezed toward the y-axis: horizontal compression by a factor of 5
    2. It is pulled away from the y-axis: horizontal stretch by a factor of 5
    3. It is stretched vertically by a factor of 5
  3. How does the graph of y=f(3x)y = f(3x) compare with the graph of y=f(x)y = f(x)?
    1. It is squeezed toward the y-axis: horizontal compression by a factor of 3
    2. It is pulled away from the y-axis: horizontal stretch by a factor of 3
    3. It is stretched vertically by a factor of 3

Watch out!

Practice ★★

  1. f(x)=8x3−9xf(x) = 8x^3 - 9x. Is ff even, odd, or neither?
    1. Even
    2. Odd
    3. Neither
  2. f(x)=9x3−2xf(x) = 9x^3 - 2x. Is ff even, odd, or neither?
    1. Even
    2. Odd
    3. Neither
  1. f(x)=3x4−8x2+7f(x) = 3x^4 - 8x^2 + 7. Is ff even, odd, or neither?
    1. Even
    2. Odd
    3. Neither
  2. f(x)=5x4−1x2+9f(x) = 5x^4 - 1x^2 + 9. Is ff even, odd, or neither?
    1. Even
    2. Odd
    3. Neither

Exit ticket

Three questions to check you have got it.

  1. f(x)=8x3−9xf(x) = 8x^3 - 9x. Is ff even, odd, or neither?
    1. Even
    2. Odd
    3. Neither
  2. f(x)=9x3−2xf(x) = 9x^3 - 2x. Is ff even, odd, or neither?
    1. Even
    2. Odd
    3. Neither
  3. f(x)=7x3−1xf(x) = 7x^3 - 1x. Is ff even, odd, or neither?
    1. Even
    2. Odd
    3. Neither

Summary card

Stretches, even and odd

  • af(x)a f(x): vertical stretch by aa. f(bx)f(bx): horizontal compression by bb.
  • Even: f(−x)=f(x)f(-x) = f(x), mirror image in the yy-axis.
  • Odd: f(−x)=−f(x)f(-x) = -f(x), symmetric about the origin.

Watch out: compute f(−x)f(-x); the signs of the coefficients do not decide.

4. Inverse functions

Warm-up

Three quick questions from earlier lessons.

Discover

f(x)=x+7f(x) = x + 7 adds 7. Which function undoes it, and gives back the number you started with?

  1. x−7x - 7
  2. 1x+7\dfrac{1}{x + 7}
  3. x+7x + 7

Key idea

The inverse function f−1f^{-1} undoes ff: it swaps inputs and outputs.

If f(a)=bf(a) = b, then f−1(b)=af^{-1}(b) = a. To find f−1(x)f^{-1}(x): write y=f(x)y = f(x), solve for xx, then rename. The graph of f−1f^{-1} is the reflection of the graph of ff across the line y=xy = x.

f−1(x)f^{-1}(x) is not 1f(x)\dfrac{1}{f(x)}: the −1-1 is not an exponent here.

Worked example

f(x)=3x−5f(x) = 3x - 5. Find f−1(x)f^{-1}(x).

y=3x−5y = 3x - 5
Name the output y.
y+5=3xy + 5 = 3x
f subtracts 5 last, so undo that first: add 5.
x=y+53x = \dfrac{y + 5}{3}
Then undo × 3: divide by 3.
f−1(x)=x+53f^{-1}(x) = \dfrac{x + 5}{3}
Rename the input x.

Your turn, guided

Your turn: f(x)=2x+6f(x) = 2x + 6.

y=2x+6y = 2x + 6
Name the output y.
y−6=2xy - 6 = 2x
Undo + 6 first.
f−1(x)=x−62f^{-1}(x) = \dfrac{x - 6}{2}
Divide by 2, then rename.

Practice ★

  1. f(x)=7x+9f(x) = 7x + 9. Find f−1(x)f^{-1}(x).
  2. f(x)=2x+2f(x) = 2x + 2. Find f−1(x)f^{-1}(x).
  3. f(x)=7x+5f(x) = 7x + 5. Find f−1(x)f^{-1}(x).
  4. f(x)=2x+9f(x) = 2x + 9. Find f−1(x)f^{-1}(x).

Watch out!

Practice ★★

  1. f(x)=2x+6f(x) = 2x + 6. Find f−1(10)f^{-1}(10).
  2. f(x)=2x+9f(x) = 2x + 9. Find f−1(11)f^{-1}(11).
  3. f(x)=3x+1f(x) = 3x + 1. Find f−1(10)f^{-1}(10).
  1. The point (2,7)(2, 7) is on the graph of ff. Tap the point that must be on the graph of f−1f^{-1}.
  2. The point (5,7)(5, 7) is on the graph of ff. Tap the point that must be on the graph of f−1f^{-1}.

Exit ticket

Three questions to check you have got it.

  1. f(x)=7x+9f(x) = 7x + 9. Find f−1(x)f^{-1}(x).
  2. f(x)=2x+2f(x) = 2x + 2. Find f−1(x)f^{-1}(x).
  3. f(x)=7x+5f(x) = 7x + 5. Find f−1(x)f^{-1}(x).

Summary card

Inverse functions

  1. Write y=f(x)y = f(x).
  2. Solve for xx, undoing the last operation first.
  3. Rename: f−1(x)f^{-1}(x). Check that f(f−1(x))=xf(f^{-1}(x)) = x.

Watch out: f−1(x)f^{-1}(x) is not 1f(x)\dfrac{1}{f(x)}.

5. Asymptotes and shifts of exponential and logarithmic graphs

Warm-up

Three quick questions from earlier lessons.

Discover

Move the sliders and watch the graph of y=a⋅2x+ky = a \cdot 2^x + k.

The graph of y = a·2^x + k, which changes as you move the sliders for a and k.

What does kk change?

  1. It moves the graph and its horizontal asymptote up or down
  2. It only moves the point where the graph crosses the y-axis
  3. It makes the graph steeper

Key idea

Shifts move the asymptotes with the graph.

y=a⋅bx+ky = a \cdot b^{x} + k has the horizontal asymptote y=ky = k. y=log⁡b(x−h)+ky = \log_b(x - h) + k has the vertical asymptote x=hx = h and domain x>hx > h

A vertical stretch by aa does not move a horizontal asymptote.

Worked example

Describe y=3⋅2x+4y = 3 \cdot 2^x + 4.

Horizontal asymptote: y=4y = 4
As x → −∞, 3 · 2^x → 0.
yy-intercept: 3+4=73 + 4 = 7
At x = 0, 2⁰ = 1.

Your turn, guided

Your turn: y=log⁡3(x−2)+5y = \log_3(x - 2) + 5. Fill in the missing line.

Domain: x>2x > 2
The argument x − 2 must be positive.
Vertical asymptote: x=2x = 2
The asymptote is where the argument equals 0.

Practice ★

  1. The graph of y=3⋅4x+7y = 3\cdot 4^x + 7 has a horizontal asymptote y=cy = c. What is cc?
  2. The graph of y=8⋅5x+5y = 8\cdot 5^x + 5 has a horizontal asymptote y=cy = c. What is cc?
  3. The graph of y=2⋅3x+9y = 2\cdot 3^x + 9 has a horizontal asymptote y=cy = c. What is cc?
  4. The graph of y=6⋅5x+2y = 6\cdot 5^x + 2 has a horizontal asymptote y=cy = c. What is cc?
  5. The graph of y=7⋅2x+2y = 7\cdot 2^x + 2 has a horizontal asymptote y=cy = c. What is cc?

Watch out!

Practice ★★

  1. Draw the graph of y=1⋅2x+3y = 1 \cdot 2^x + 3: place two points and the asymptote.
  2. Draw the graph of y=1⋅2x+1y = 1 \cdot 2^x + 1: place two points and the asymptote.
  1. The graph of y=log⁡4(x−6)+1y = \log_{ 4 }(x - 6) + 1 has a vertical asymptote x=cx = c. What is cc?
  2. The graph of y=log⁡2(x−4)+8y = \log_{ 2 }(x - 4) + 8 has a vertical asymptote x=cx = c. What is cc?
  3. The graph of y=log⁡2(x−9)+3y = \log_{ 2 }(x - 9) + 3 has a vertical asymptote x=cx = c. What is cc?
  4. The graph of y=log⁡4(x−9)+5y = \log_{ 4 }(x - 9) + 5 has a vertical asymptote x=cx = c. What is cc?
  5. The graph of y=log⁡4(x−5)+3y = \log_{ 4 }(x - 5) + 3 has a vertical asymptote x=cx = c. What is cc?

Exit ticket

Three questions to check you have got it.

  1. The graph of y=log⁡4(x−6)+1y = \log_{ 4 }(x - 6) + 1 has a vertical asymptote x=cx = c. What is cc?
  2. The graph of y=log⁡2(x−4)+8y = \log_{ 2 }(x - 4) + 8 has a vertical asymptote x=cx = c. What is cc?
  3. The graph of y=log⁡2(x−9)+3y = \log_{ 2 }(x - 9) + 3 has a vertical asymptote x=cx = c. What is cc?

Summary card

Asymptotes after shifts

  • y=a⋅bx+ky = a \cdot b^{x} + k: horizontal asymptote y=ky = k.
  • y=log⁡b(x−h)+ky = \log_b(x - h) + k: vertical asymptote x=hx = h, domain x>hx > h.