AP Precalculus · Lesson 3

Transformations and equivalent forms

Know Know it

For g(x)=a f(b(x−h))+kg(x)=a\,f(b(x-h))+k: kk shifts the graph up (k>0k>0) or down, and hh shifts it right (h>0h>0) or left. The factor aa is a vertical dilation by ∣a∣|a|, with a reflection over the xx-axis when a<0a < 0. The factor bb is a horizontal dilation by 1∣b∣\frac{1}{|b|}, with a reflection over the yy-axis when b<0b < 0.

Track one point to check your work. If (4,6)(4,6) is on ff, then on f(x−2)+1f(x-2)+1 it moves to (6,7)(6,7).

Different but equivalent forms show different features. Factored form shows the zeros. Standard form shows the degree, the leading term and the yy-intercept. For a rational function, polynomial long division writes it as a quotient plus a remainder, and the quotient gives the slant or horizontal asymptote.

The Binomial Theorem expands (a+b)n(a+b)^n without multiplying it out: the coefficients come from row nn of Pascal's triangle. For example, row 4 is 1, 4, 6, 4, 1.

Trap Inside changes go the other way

In f(x−3)f(x-3) the graph moves 3 units to the right, not left, and in f(2x)f(2x) it is squeezed toward the yy-axis by a factor of 12\frac{1}{2}, not stretched. Changes inside the parentheses act on inputs, so they undo what the numbers seem to say.

Worked example Easy

The point (4,6)(4,6) is on the graph of ff. Let g(x)=f(x−2)+1g(x)=f(x-2)+1. Which point must be on the graph of gg?

  1. (2,7)(2,7)
  2. (6,5)(6,5)
  3. (2,5)(2,5)
  4. (6,7)(6,7)Answer

How to solve it

  1. Choose xx so the input to ff is 4: x−2=4x-2=4 gives x=6x=6.
  2. Then g(6)=f(4)+1=6+1=7g(6)=f(4)+1=6+1=7.
  3. The point is (6,7)(6,7): right 2 and up 1.

Why each choice is right or wrong

  • A. Incorrect. The point (2,7)(2,7) moves the graph left 2, but x−2x-2 inside moves it right 2.
  • B. Incorrect. The point (6,5)(6,5) shifts down 1, but adding 1 outside raises every output by 1.
  • C. Incorrect. The point (2,5)(2,5) reverses both shifts: it should go right 2 and up 1.
  • D. Correct. g(6)=f(6−2)+1=f(4)+1=7g(6)=f(6-2)+1=f(4)+1=7, so (6,7)(6,7) is on gg.

Explain Explain it

Task verbs: describe, calculate

To describe a transformation, name each change, its size and its direction, in an order that works: dilations and reflections first, then shifts.

Worked example Medium

Let r(x)=2x2+3x−4x+2r(x)=\frac{2x^2+3x-4}{x+2}. What is the equation of the slant asymptote of the graph of rr?

  1. y=2x−1y=2x-1Answer
  2. y=2x+7y=2x+7
  3. y=2xy=2x
  4. y=2x+3y=2x+3

How to solve it

  1. Divide: 2x2÷x=2x2x^2 \div x = 2x, and 2x(x+2)=2x2+4x2x(x+2)=2x^2+4x; subtracting leaves −x−4-x-4.
  2. Next, −x÷x=−1-x \div x=-1, and −1(x+2)=−x−2-1(x+2)=-x-2; subtracting leaves −2-2.
  3. So r(x)=2x−1−2x+2r(x)=2x-1-\frac{2}{x+2}, and the slant asymptote is y=2x−1y=2x-1.

Why each choice is right or wrong

  • A. Correct. Division gives r(x)=2x−1−2x+2r(x)=2x-1-\frac{2}{x+2}. The remainder part goes to 0 for large xx, so the graph approaches y=2x−1y=2x-1.
  • B. Incorrect. The line y=2x+7y=2x+7 comes from dividing as if by x−2x-2, using 2 in place of -2.
  • C. Incorrect. The line y=2xy=2x keeps only the ratio of leading terms and drops the constant part of the quotient.
  • D. Incorrect. The line y=2x+3y=2x+3 comes from dividing 2x2+3x2x^2+3x by xx alone, dropping the −4-4 and the +2+2 in the denominator, which is not a valid division.

Describe Your turn

The function ff has a maximum value of 5. Let g(x)=−2f(x)+1g(x)=-2f(x)+1. Describe the transformations that take the graph of ff to the graph of gg, and give the minimum value of gg.

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Compare with a model answer

A full answer does this

  • Names a vertical dilation by a factor of 2 and a reflection over the xx-axis
  • Names a shift up of 1 unit
  • Gives the minimum value of gg as -9

Model answer

Multiplying by -2 stretches the graph of ff vertically by a factor of 2 and reflects it over the xx-axis. Adding 1 then shifts it up 1 unit. The reflection turns the maximum of ff into the minimum of gg, so the minimum value of gg is −2(5)+1=−9-2(5)+1=-9.

Apply Apply it

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