Question

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

y = –2x^2 + 32x –12

Answer 1

find f'(x) and equate it to 0
solve for x

Answer 2

I don't understand that ?

Answer 3

oh - i've assumed you know some calculus

Answer 4

Would it be like y=-2(0)^2+32(0)-12 ?

Answer 5

no what you have there is the intercept on y-axis

Answer 6

you need to change it to the vertex form
y = –2x^2 + 32x –12
= -2(x^2 - 16x + 6)
= -2( x - 8)^2 -64 + 6)
= -2[(x-8)^2 - 58)]
so the x coordinate at maximum = 8
a negative coefficient before x^3 means there is a maximum value

Answer 7

oh , but it says the maximum and range is only 116 . its either 116 or -116 ?

Answer 8

at maximum the function has a value -2*-58 = 116

Answer 9

ok , i see that . . .

Answer 10

the range is f(x)<= 116

Answer 11

less than or greater than ?