## andreascott 3 years ago What is the maximum or minimum value of the function? What is the range? y = –2x^2 + 32x –12

1. cwrw238

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

2. andreascott

I don't understand that ?

3. cwrw238

oh - i've assumed you know some calculus

4. andreascott

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

5. cwrw238

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

6. cwrw238

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

7. andreascott

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

8. cwrw238

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

9. andreascott

ok , i see that . . .

10. cwrw238

the range is f(x)<= 116

11. andreascott

less than or greater than ?

12. cwrw238

|dw:1350681832911:dw|

13. cwrw238

below the graph less or equal to 116

14. andreascott

Thats what i thought it was (: . Thank you soo much for helping me ! (;

15. cwrw238

yw