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=3f(x)−2y = 3f(x) - 2y=3f(x)−2?
Solution
Correct answer: D
The output is scaled by 333, then lowered by 222; the input does not change.
(a, 3b−2)(a,\; 3b - 2)(a,3b−2)
A point (p,q)(p, q)(p,q) lies on the graph of y=f(x)y = f(x)y=f(x). Where does it move on the graph of y=f(3x)+4y = f(3x) + 4y=f(3x)+4?
Correct answer: B
The inside 3x3x3x reaches the old input at one third of the xxx-value, and the +4+4+4 raises the output.
(p3, q+4)\left(\tfrac{p}{3},\; q + 4\right)(3p,q+4)
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) stretched vertically by 555 and reflected across the x-axis. If f(−1)=2f(-1) = 2f(−1)=2, what is g(−1)g(-1)g(−1)?
Correct answer: C
Stretching and reflecting the output means g(x)=−5f(x)g(x) = -5f(x)g(x)=−5f(x).
g(−1)=−5×f(−1)=−5×2=−10g(-1) = -5 \times f(-1) = -5 \times 2 = -10g(−1)=−5×f(−1)=−5×2=−10
The graph of y=f(x)y = f(x)y=f(x) has domain −6≤x≤3-6 \le x \le 3−6≤x≤3 and range −2≤y≤10-2 \le y \le 10−2≤y≤10. What are the domain and range of y=f(−x)y = f(-x)y=f(−x)?
Reflecting across the y-axis negates every input, so the domain endpoints change sign and swap, while the range is untouched.
−6≤x≤3 ⟶ −3≤x≤6-6 \le x \le 3 \;\longrightarrow\; -3 \le x \le 6−6≤x≤3⟶−3≤x≤6
The point (−4,−8)(-4, -8)(−4,−8) lies on the graph of y=f(x)y = f(x)y=f(x). Which point must lie on the graph of y=−14f(x)y = -\tfrac{1}{4}f(x)y=−41f(x)?
The factor −14-\tfrac{1}{4}−41 multiplies the height, scaling and flipping it; the input stays.
(−4, −14×(−8))=(−4,2)\left(-4,\; -\tfrac{1}{4} \times (-8)\right) = (-4, 2)(−4,−41×(−8))=(−4,2)
A parabola with vertex (0,0)(0, 0)(0,0) passes through (2,−2)(2, -2)(2,−2) and opens downward. Which equation is it?
Correct answer: A
With vertex at the origin the equation is y=ax2y = ax^2y=ax2. Use the point (2,−2)(2, -2)(2,−2).
−2=a(2)2=4a ⇒ a=−12-2 = a(2)^2 = 4a \;\Rightarrow\; a = -\tfrac{1}{2}−2=a(2)2=4a⇒a=−21
The negative value opens the bowl downward, so y=−12x2y = -\tfrac{1}{2}x^2y=−21x2.
The graph of y=f(x)y = f(x)y=f(x) has an x-intercept at (5,0)(5, 0)(5,0) and a y-intercept at (0,7)(0, 7)(0,7). After reflecting the graph across the y-axis, where is the y-intercept?
The y-intercept sits at input 000, and reflecting across the y-axis negates the input.
−0=0 ⇒ the y-intercept stays at (0,7)-0 = 0 \;\Rightarrow\; \text{the y-intercept stays at } (0, 7)−0=0⇒the y-intercept stays at (0,7)
Only points with a nonzero input change sides.
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) stretched vertically by 222, reflected across the x-axis, and shifted up 555, so g(x)=−2f(x)+5g(x) = -2f(x) + 5g(x)=−2f(x)+5. If f(4)=3f(4) = 3f(4)=3, what is g(4)g(4)g(4)?
Substitute f(4)=3f(4) = 3f(4)=3 into the rule for ggg.
g(4)=−2×3+5=−6+5=−1g(4) = -2 \times 3 + 5 = -6 + 5 = -1g(4)=−2×3+5=−6+5=−1
A function fff has domain 2≤x≤82 \le x \le 82≤x≤8. What is the domain of y=f(x2)y = f\left(\tfrac{x}{2}\right)y=f(2x)?
The inside x2\tfrac{x}{2}2x reaches each old input at twice the xxx-value, so the domain endpoints double.
2≤x≤8 ⟶ 4≤x≤162 \le x \le 8 \;\longrightarrow\; 4 \le x \le 162≤x≤8⟶4≤x≤16
The graph of y=f(x)y = f(x)y=f(x) passes through the origin (0,0)(0, 0)(0,0). Which is guaranteed to keep the graph passing through the origin: a reflection across the y-axis, a vertical stretch, both, or neither?
The origin has input 000 and height 000. A reflection across the y-axis sends it to (−0,0)=(0,0)(-0, 0) = (0, 0)(−0,0)=(0,0), and a vertical stretch sends it to (0, a×0)=(0,0)(0,\, a \times 0) = (0, 0)(0,a×0)=(0,0).
(0,0)⟶(0,0)(0, 0) \longrightarrow (0, 0)(0,0)⟶(0,0)
So both transformations fix the origin.
The graph of y=cf(x)y = cf(x)y=cf(x) passes through (2,−6)(2, -6)(2,−6), and f(2)=3f(2) = 3f(2)=3. What is ccc?
At x=2x = 2x=2 the scaled height is c⋅f(2)c \cdot f(2)c⋅f(2), and that equals −6-6−6.
c×3=−6 ⇒ c=−2c \times 3 = -6 \;\Rightarrow\; c = -2c×3=−6⇒c=−2
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)?
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 the input −a-a−a produces height bbb on the original.
(−a, b) lies on y=f(x)(-a,\; b) \text{ lies on } y = f(x)(−a,b) lies on y=f(x)
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