Question

What is the maximum or minimum value of the function? What is the range?

y = –2x2 + 32x –12 (1 point)
maximum: 116
range: y 116
maximum: –116
range: y –116
maximum: 116
range: y 116
maximum: –116
range: y –116

Answer 1

\[-2x^2+32x-12 = -2(x^2-16x+6)\]
completing the square gives:
\[-2(x^2-16x +8^2-(8^2) +6)\]
\[-2((x-8)^2-(64) +6)\]
\[-2((x-8)^2-58)=-2(x-8)^2+116\]
Therefore the max value is at point (8,116)

Answer 2

so it b

Answer 3

C

Answer 4

o ok can u help me with another one

Answer 5

Well I don't understand you multiple choices. The function has a maximum of 116
and a range of
\[(-\infty, 116]\]

Answer 6

What is the graph of the equation?

y = x2 – 4x + 3

Answer 7

You can factor this to
y = (x-3)(x-1)
so it is a parabola, open up. You know this because the squared term is positive. If it were negative it would open down.
It goes through the points (3,0) and (1,0).
The vertex you will need to use the formula or complete the square to get to.
I think that the vertex form is y=(x-2)^2 -1 so the vertex would be at (2,-1).

Answer 8

4. What is the maximum or minimum value of the function? What is the range?

y = –2x2 + 32x –12
(1 point)
(1 pt) maximum: 116
range: y 116

(0 pts) maximum: –116
range: y –116

(0 pts) maximum: 116
range: y 116

(0 pts) maximum: –116
range: y –116

1 /1 point