12 multiple-choice questions, progressively harder.
A parabola has vertex (−1,4)(-1, 4)(−1,4) and opens downward. What is its range?
Solution
Correct answer: C
Opening downward, the vertex is the highest point, so the vertex height is the largest output.
y≤4y \le 4y≤4
The outputs run from 444 downward, so the range is (−∞,4](-\infty, 4](−∞,4].
What is the vertex of f(x)=2x2+4x+5f(x) = 2x^2 + 4x + 5f(x)=2x2+4x+5?
Correct answer: B
With a=2a = 2a=2 and b=4b = 4b=4, the axis is x=−44=−1x = -\dfrac{4}{4} = -1x=−44=−1.
f(−1)=2−4+5=3f(-1) = 2 - 4 + 5 = 3f(−1)=2−4+5=3
So the vertex is (−1,3)(-1, 3)(−1,3).
A quadratic satisfies f(2)=f(−6)f(2) = f(-6)f(2)=f(−6). What is its axis of symmetry?
Equal outputs make 222 and −6-6−6 mirror partners, so the axis is their midpoint.
x=2+(−6)2=−42=−2x = \frac{2 + (-6)}{2} = \frac{-4}{2} = -2x=22+(−6)=2−4=−2
The axis of symmetry is x=−2x = -2x=−2.
Compared with the parent y=x2y = x^2y=x2, the graph of y=12x2y = \tfrac{1}{2}x^2y=21x2 is
Correct answer: A
The factor 12\tfrac{1}{2}21 multiplies every height, so the arms climb half as fast as the parent's.
∣a∣=12<1 ⇒ vertical compression|a| = \tfrac{1}{2} < 1 \ \Rightarrow\ \text{vertical compression}∣a∣=21<1 ⇒ vertical compression
Slower-rising arms make the parabola wider than the parent.
A parabola has vertex value f(m)=4f(m) = 4f(m)=4 and leading coefficient a=−1a = -1a=−1. Using f(m+t)=f(m)+at2f(m + t) = f(m) + at^2f(m+t)=f(m)+at2, what is the output 333 units from the axis?
Correct answer: D
Substitute t=3t = 3t=3 into the displacement identity, keeping the sign of aaa.
f(m+3)=4+(−1)(3)2=4−9=−5f(m + 3) = 4 + (-1)(3)^2 = 4 - 9 = -5f(m+3)=4+(−1)(3)2=4−9=−5
The output 333 units from the axis is −5-5−5.
The function f(x)=x2−10x+21f(x) = x^2 - 10x + 21f(x)=x2−10x+21 factors as (x−3)(x−7)(x - 3)(x - 7)(x−3)(x−7). What is its minimum value?
The zeros 333 and 777 put the axis at their midpoint x=5x = 5x=5, and the parabola opens upward.
f(5)=25−50+21=−4f(5) = 25 - 50 + 21 = -4f(5)=25−50+21=−4
The minimum value is −4-4−4.
A quadratic has zeros −2-2−2 and 666 and passes through (2,−16)(2, -16)(2,−16). Written as a(x+2)(x−6)a(x + 2)(x - 6)a(x+2)(x−6), what is aaa?
Substitute the point (2,−16)(2, -16)(2,−16) into the factored form.
−16=a(2+2)(2−6)=a(4)(−4)=−16a ⇒ a=1-16 = a(2 + 2)(2 - 6) = a(4)(-4) = -16a \ \Rightarrow\ a = 1−16=a(2+2)(2−6)=a(4)(−4)=−16a ⇒ a=1
So a=1a = 1a=1.
What is the axis of symmetry of f(x)=4x2−20x+9f(x) = 4x^2 - 20x + 9f(x)=4x2−20x+9?
With a=4a = 4a=4 and b=−20b = -20b=−20, apply x=−b2ax = -\dfrac{b}{2a}x=−2ab.
x=−−202⋅4=208=52x = -\frac{-20}{2 \cdot 4} = \frac{20}{8} = \frac{5}{2}x=−2⋅4−20=820=25
The axis of symmetry is x=52x = \dfrac{5}{2}x=25.
A downward-opening parabola has vertex (3,0)(3, 0)(3,0). How many xxx-intercepts does it have?
The vertex (3,0)(3, 0)(3,0) sits on the xxx-axis, and opening downward every other point is below it.
y≤0,y=0 only at x=3y \le 0, \quad y = 0 \text{ only at } x = 3y≤0,y=0 only at x=3
The graph meets the axis at exactly one point, the vertex itself.
For an upward-opening parabola with two distinct zeros, the outputs between the zeros are
The parabola touches the axis at the two zeros and dips to its vertex between them; opening upward, that vertex is below the axis.
r1<x<r2 ⇒ f(x)<0r_1 < x < r_2 \ \Rightarrow\ f(x) < 0r1<x<r2 ⇒ f(x)<0
So the outputs between the zeros are negative.
The graph of f(x)=ax2+bx+cf(x) = ax^2 + bx + cf(x)=ax2+bx+c passes through (0,−4)(0, -4)(0,−4). Which coefficient is thereby determined?
Passing through (0,−4)(0, -4)(0,−4) means f(0)=−4f(0) = -4f(0)=−4, and f(0)f(0)f(0) is the constant term.
f(0)=c=−4f(0) = c = -4f(0)=c=−4
So c=−4c = -4c=−4 is fixed, while aaa and bbb remain free.
The parent y=x2y = x^2y=x2 is transformed into y=(x−5)2−2y = (x - 5)^2 - 2y=(x−5)2−2. What is the new vertex?
Match a(x−h)2+ka(x - h)^2 + ka(x−h)2+k: the inside gives h=5h = 5h=5 and the outside gives k=−2k = -2k=−2.
vertex (h,k)=(5,−2)\text{vertex } (h, k) = (5, -2)vertex (h,k)=(5,−2)
The parent shifts right 555 and down 222, landing its vertex at (5,−2)(5, -2)(5,−2).
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