12 multiple-choice questions, progressively harder.
The point (3,−4)(3, -4)(3,−4) is on the graph of y=f(x)y = f(x)y=f(x). Which point must be on the graph of y=−2f(x)y = -2f(x)y=−2f(x)?
Solution
Correct answer: B
The factor −2-2−2 multiplies the output, scaling and flipping the height; the input is unchanged.
(3, −2×(−4))=(3,8)(3,\; -2 \times (-4)) = (3, 8)(3,−2×(−4))=(3,8)
A function fff has range −1≤y≤4-1 \le y \le 4−1≤y≤4. What is the range of y=2f(x)y = 2f(x)y=2f(x)?
Correct answer: D
A vertical stretch multiplies every output by 222, so both ends of the range double.
−1≤y≤4 ⟶ −2≤y≤8-1 \le y \le 4 \;\longrightarrow\; -2 \le y \le 8−1≤y≤4⟶−2≤y≤8
A function fff has domain 0≤x≤60 \le x \le 60≤x≤6. What is the domain of y=f(3x)y = f(3x)y=f(3x)?
The inside 3x3x3x reaches each old input at one third of the xxx-value, so the domain endpoints divide by 333.
0≤x≤6 ⟶ 0≤x≤20 \le x \le 6 \;\longrightarrow\; 0 \le x \le 20≤x≤6⟶0≤x≤2
The point (2,7)(2, 7)(2,7) is on the graph of y=f(x)y = f(x)y=f(x). Which point must be on the graph of y=f(−x)y = f(-x)y=f(−x)?
Reflecting across the y-axis negates the input and keeps the height.
(2, 7)⟶(−2,7)(2,\; 7) \longrightarrow (-2, 7)(2,7)⟶(−2,7)
How is the graph of y=f(x2)y = f\left(\tfrac{x}{2}\right)y=f(2x) related to the graph of y=f(x)y = f(x)y=f(x)?
Correct answer: C
The inside x2\tfrac{x}{2}2x means b=12b = \tfrac{1}{2}b=21, and each input scales by 1b=2\tfrac{1}{b} = 2b1=2, pushing points away from the y-axis.
y=f(x2) ⇒ horizontal stretch by 2y = f\left(\tfrac{x}{2}\right) \;\Rightarrow\; \text{horizontal stretch by } 2y=f(2x)⇒horizontal stretch by 2
The point (5,−3)(5, -3)(5,−3) is on the graph of y=f(x)y = f(x)y=f(x). Which point must be on the graph of y=−f(x)y = -f(x)y=−f(x)?
Negating the output flips the height's sign while the input stays put.
(5, −(−3))=(5,3)(5,\; -(-3)) = (5, 3)(5,−(−3))=(5,3)
Which equation compresses y=x2y = x^2y=x2 vertically to one third of its height?
Correct answer: A
A vertical compression multiplies each output by a factor between 000 and 111. One third of each height means multiply by 13\tfrac{1}{3}31.
y=13x2y = \tfrac{1}{3}x^2y=31x2
A function fff has domain 1≤x≤51 \le x \le 51≤x≤5. What is the domain of y=f(−x)y = f(-x)y=f(−x)?
Reflecting across the y-axis negates every input, so each endpoint changes sign and the two ends swap order.
1≤x≤5 ⟶ −5≤x≤−11 \le x \le 5 \;\longrightarrow\; -5 \le x \le -11≤x≤5⟶−5≤x≤−1
The point (−2,6)(-2, 6)(−2,6) is on the graph of y=f(x)y = f(x)y=f(x). Which point must be on the graph of y=13f(x)y = \tfrac{1}{3}f(x)y=31f(x)?
The factor 13\tfrac{1}{3}31 multiplies the height and leaves the input unchanged.
(−2, 13×6)=(−2,2)\left(-2,\; \tfrac{1}{3} \times 6\right) = (-2, 2)(−2,31×6)=(−2,2)
The graph of y=f(x)y = f(x)y=f(x) is compressed horizontally toward the y-axis by a factor of 12\tfrac{1}{2}21. Which equation gives it?
A horizontal compression by 12\tfrac{1}{2}21 scales each input by 12\tfrac{1}{2}21, which needs b=2b = 2b=2 inside the function.
y=f(2x)y = f(2x)y=f(2x)
Dividing the input would stretch it, and factors outside change the height instead.
What is the vertex of the parabola y=−12(x+4)2−3y = -\tfrac{1}{2}(x + 4)^2 - 3y=−21(x+4)2−3?
Read x+4x + 4x+4 as x−(−4)x - (-4)x−(−4), so h=−4h = -4h=−4, and the constant gives k=−3k = -3k=−3; the factor −12-\tfrac{1}{2}−21 affects only the width and opening.
vertex=(−4,−3)\text{vertex} = (-4, -3)vertex=(−4,−3)
A function fff has range 2≤y≤102 \le y \le 102≤y≤10. What is the range of y=12f(x)y = \tfrac{1}{2}f(x)y=21f(x)?
A vertical compression multiplies every output by 12\tfrac{1}{2}21, so both ends of the range halve.
2≤y≤10 ⟶ 1≤y≤52 \le y \le 10 \;\longrightarrow\; 1 \le y \le 52≤y≤10⟶1≤y≤5
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