12 multiple-choice questions, progressively harder.
The point (a,b)(a, b)(a,b) lies on the graph of y=f(x)y = f(x)y=f(x). Which point must lie on the graph of y=−2f(x)+1y = -2f(x) + 1y=−2f(x)+1?
Solution
Correct answer: A
The output is scaled by −2-2−2 (stretch and flip), then raised by 111; the input is untouched.
(a, −2b+1)(a,\; -2b + 1)(a,−2b+1)
The graph of y=f(x)y = f(x)y=f(x) passes through (4,3)(4, 3)(4,3). Which point is guaranteed to lie on the graph of y=12f(x−2)y = \tfrac{1}{2}f(x - 2)y=21f(x−2)?
Correct answer: C
The inside x−2x - 2x−2 shifts the point right 222, and the factor 12\tfrac{1}{2}21 halves its height.
(4+2, 12×3)=(6,32)\left(4 + 2,\; \tfrac{1}{2} \times 3\right) = \left(6, \tfrac{3}{2}\right)(4+2,21×3)=(6,23)
The graph of y=g(x)y = g(x)y=g(x) is the graph of y=f(x)y = f(x)y=f(x) reflected across the x-axis and stretched vertically by 333. If f(2)=5f(2) = 5f(2)=5, what is g(2)g(2)g(2)?
Correct answer: D
Reflecting and stretching the output means g(x)=−3f(x)g(x) = -3f(x)g(x)=−3f(x).
g(2)=−3×f(2)=−3×5=−15g(2) = -3 \times f(2) = -3 \times 5 = -15g(2)=−3×f(2)=−3×5=−15
The parabola y=−2(x−1)2+8y = -2(x - 1)^2 + 8y=−2(x−1)2+8 has vertex (1,8)(1, 8)(1,8) and opens downward. Where does it cross the x-axis?
Set the output to 000 and solve for xxx.
−2(x−1)2+8=0 ⇒ (x−1)2=4 ⇒ x=3 or x=−1-2(x - 1)^2 + 8 = 0 \;\Rightarrow\; (x - 1)^2 = 4 \;\Rightarrow\; x = 3 \text{ or } x = -1−2(x−1)2+8=0⇒(x−1)2=4⇒x=3 or x=−1
The crossings are (3,0)(3, 0)(3,0) and (−1,0)(-1, 0)(−1,0), symmetric about the vertex input x=1x = 1x=1.
The graph of y=f(x)y = f(x)y=f(x) has domain −2≤x≤6-2 \le x \le 6−2≤x≤6 and range 0≤y≤120 \le y \le 120≤y≤12. What are the domain and range of y=13f(x)y = \tfrac{1}{3}f(x)y=31f(x)?
A vertical scaling touches only the range. The factor 13\tfrac{1}{3}31 multiplies each output, so the range shrinks while the domain is unchanged.
0≤y≤12 ⟶ 0≤y≤40 \le y \le 12 \;\longrightarrow\; 0 \le y \le 40≤y≤12⟶0≤y≤4
A point (a,b)(a, b)(a,b) lies on the graph of y=−f(x)y = -f(x)y=−f(x). Which point must lie on the original graph y=f(x)y = f(x)y=f(x)?
Correct answer: B
If (a,b)(a, b)(a,b) is on y=−f(x)y = -f(x)y=−f(x) then b=−f(a)b = -f(a)b=−f(a), so f(a)=−bf(a) = -bf(a)=−b.
(a, −b) lies on y=f(x)(a,\; -b) \text{ lies on } y = f(x)(a,−b) lies on y=f(x)
Undoing an x-axis reflection just negates the height back.
Which transformation turns y=x2y = x^2y=x2 into y=−14x2y = -\tfrac{1}{4}x^2y=−41x2?
Write the coefficient as a=−14a = -\tfrac{1}{4}a=−41. Its size 14\tfrac{1}{4}41 is a vertical compression, and its negative sign reflects across the x-axis.
a=−14 ⇒ reflect across x-axis, compress to 14a = -\tfrac{1}{4} \;\Rightarrow\; \text{reflect across x-axis, compress to } \tfrac{1}{4}a=−41⇒reflect across x-axis, compress to 41
The point (−6,9)(-6, 9)(−6,9) lies on the graph of y=f(x)y = f(x)y=f(x). Which point must lie on the graph of y=f(x3)y = f\left(\tfrac{x}{3}\right)y=f(3x)?
A point reappears where the inside x3\tfrac{x}{3}3x equals the old input −6-6−6.
x3=−6 ⇒ x=−18\tfrac{x}{3} = -6 \;\Rightarrow\; x = -183x=−6⇒x=−18
The height stays 999, so the point moves to (−18,9)(-18, 9)(−18,9).
Which equation is NOT the same width as y=x2y = x^2y=x2?
Width depends only on the size of the coefficient on the squared term. Three of these have size 111, but y=2x2y = 2x^2y=2x2 has size 222.
∣2∣=2≠1 ⇒ narrower than y=x2|2| = 2 \ne 1 \;\Rightarrow\; \text{narrower than } y = x^2∣2∣=2=1⇒narrower than y=x2
A minus sign or an added constant does not change the width.
The point (2,−6)(2, -6)(2,−6) lies on the graph of y=f(x)y = f(x)y=f(x). Which point must lie on the graph of y=−12f(x)y = -\tfrac{1}{2}f(x)y=−21f(x)?
The factor −12-\tfrac{1}{2}−21 multiplies the height, halving it and flipping its sign; the input stays.
(2, −12×(−6))=(2,3)\left(2,\; -\tfrac{1}{2} \times (-6)\right) = (2, 3)(2,−21×(−6))=(2,3)
Which single equation reflects the graph of y=f(x)y = f(x)y=f(x) across the y-axis and stretches it vertically by 444?
Reflecting across the y-axis negates the input, giving f(−x)f(-x)f(−x), and stretching vertically by 444 multiplies the output by 444.
y=4f(−x)y = 4f(-x)y=4f(−x)
The parabola y=ax2y = ax^2y=ax2 passes through (2,2)(2, 2)(2,2). What is aaa, and is the bowl wider or narrower than y=x2y = x^2y=x2?
Substitute the point to find aaa.
2=a(2)2=4a ⇒ a=122 = a(2)^2 = 4a \;\Rightarrow\; a = \tfrac{1}{2}2=a(2)2=4a⇒a=21
Since the size 12\tfrac{1}{2}21 is less than 111, the bowl is wider than y=x2y = x^2y=x2.
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