12 multiple-choice questions, progressively harder.
Does the quantity y=−x2+6x+1y = -x^2 + 6x + 1y=−x2+6x+1 have a maximum or a minimum?
Solution
Correct answer: C
The kind of extreme value is decided by the sign of the leading coefficient aaa, the number in front of x2x^2x2. Here a=−1a = -1a=−1.
a=−1<0 ⇒ opens downwarda = -1 < 0 \;\Rightarrow\; \text{opens downward}a=−1<0⇒opens downward
A downward parabola is a hill, so its vertex is the highest point and the quantity has a maximum.
The quantity y=2x2−8x+3y = 2x^2 - 8x + 3y=2x2−8x+3 opens upward. Its vertex is therefore a ...
Correct answer: A
An upward-opening parabola is a valley, so its turning point is the lowest point.
a=2>0 ⇒ vertex is the lowest pointa = 2 > 0 \;\Rightarrow\; \text{vertex is the lowest point}a=2>0⇒vertex is the lowest point
The lowest point is a minimum, so the quantity has a minimum there.
At what value of xxx does y=x2−10x+7y = x^2 - 10x + 7y=x2−10x+7 reach its minimum?
Correct answer: B
The minimum of an upward parabola is at the vertex, whose input is x=−b2ax = -\tfrac{b}{2a}x=−2ab with a=1a = 1a=1 and b=−10b = -10b=−10.
x=−−102(1)=102=5x = -\frac{-10}{2(1)} = \frac{10}{2} = 5x=−2(1)−10=210=5
So the minimum occurs at the input x=5x = 5x=5.
The quantity y=−3x2+12xy = -3x^2 + 12xy=−3x2+12x has a maximum. At what input does it occur?
Find the vertex input with x=−b2ax = -\tfrac{b}{2a}x=−2ab, reading a=−3a = -3a=−3 and b=12b = 12b=12.
x=−122(−3)=−12−6=2x = -\frac{12}{2(-3)} = -\frac{12}{-6} = 2x=−2(−3)12=−−612=2
The maximum occurs at the input x=2x = 2x=2.
A parabola opens upward with vertex (3,−5)(3, -5)(3,−5). What is the smallest value the quantity can take?
An upward parabola has its vertex as the lowest point, so the smallest value is the vertex's y-coordinate.
vertex (3,−5) ⇒ minimum value −5\text{vertex } (3, -5) \;\Rightarrow\; \text{minimum value } -5vertex (3,−5)⇒minimum value −5
The input 333 is where it happens; the value is −5-5−5.
A ball's height in feet is h=−16t2+96th = -16t^2 + 96th=−16t2+96t. The greatest height it reaches is found at the ... of the height parabola.
The height is a downward parabola (a=−16<0a = -16 < 0a=−16<0), so the greatest height is its highest point, the turning point.
a<0 ⇒ highest point is the vertexa < 0 \;\Rightarrow\; \text{highest point is the vertex}a<0⇒highest point is the vertex
The maximum height is the value of the quantity at the vertex.
What is the maximum value of y=−x2+8x−3y = -x^2 + 8x - 3y=−x2+8x−3?
First find the vertex input: x=−b2a=−82(−1)=4x = -\tfrac{b}{2a} = -\tfrac{8}{2(-1)} = 4x=−2ab=−2(−1)8=4. Then substitute back for the value.
y=−(4)2+8(4)−3=−16+32−3=13y = -(4)^2 + 8(4) - 3 = -16 + 32 - 3 = 13y=−(4)2+8(4)−3=−16+32−3=13
The maximum value is 131313. The input 444 is only where it occurs, not the answer.
The quantity y=x2+2x+5y = x^2 + 2x + 5y=x2+2x+5 has a minimum value. What input produces it?
The minimum is at the vertex input x=−b2ax = -\tfrac{b}{2a}x=−2ab with a=1a = 1a=1, b=2b = 2b=2.
x=−22(1)=−1x = -\frac{2}{2(1)} = -1x=−2(1)2=−1
So the minimum occurs at the input x=−1x = -1x=−1.
For y=5x2−20x+1y = 5x^2 - 20x + 1y=5x2−20x+1, the vertex is a minimum. Which fact tells you so?
Correct answer: D
Only the sign of the leading coefficient aaa decides whether the vertex is a maximum or a minimum.
a=5>0 ⇒ opens upward, so the vertex is a minimuma = 5 > 0 \;\Rightarrow\; \text{opens upward, so the vertex is a minimum}a=5>0⇒opens upward, so the vertex is a minimum
The values of bbb and ccc move the parabola around but never decide max versus min.
Find the minimum value of y=x2−2x−3y = x^2 - 2x - 3y=x2−2x−3.
The vertex input is x=−b2a=−−22=1x = -\tfrac{b}{2a} = -\tfrac{-2}{2} = 1x=−2ab=−2−2=1. Substitute back for the minimum value.
y=(1)2−2(1)−3=1−2−3=−4y = (1)^2 - 2(1) - 3 = 1 - 2 - 3 = -4y=(1)2−2(1)−3=1−2−3=−4
The minimum value is −4-4−4. The input 111 and the constant −3-3−3 are not the answer.
The quantity y=−2x2+4x+6y = -2x^2 + 4x + 6y=−2x2+4x+6 reaches its maximum at which input?
The maximum is at the vertex input x=−b2ax = -\tfrac{b}{2a}x=−2ab with a=−2a = -2a=−2, b=4b = 4b=4.
x=−42(−2)=−4−4=1x = -\frac{4}{2(-2)} = -\frac{4}{-4} = 1x=−2(−2)4=−−44=1
So the maximum occurs at the input x=1x = 1x=1.
A quantity is modeled by y=3(x−5)2+2y = 3(x - 5)^2 + 2y=3(x−5)2+2. What is its minimum value?
This is vertex form with (h,k)=(5,2)(h, k) = (5, 2)(h,k)=(5,2) and a=3>0a = 3 > 0a=3>0, so the vertex is a minimum and the minimum value is kkk.
ymin=k=2y_{\min} = k = 2ymin=k=2
The input 555 is where the minimum occurs; the minimum value is 222.
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