Please check my work to see if O am correct on minimum yvallue on the graph of y=f(x)

f(x)=-x^2+6x+8

6/-2=3
x=f(3)
(3)^2+6*(3)+8

3^2=9
6*3=18
+8
y-value is 35?

Question

Please check my work to see if O am correct on minimum yvallue on the graph of y=f(x)

f(x)=-x^2+6x+8

6/-2=3
x=f(3)
(3)^2+6*(3)+8

3^2=9
6*3=18
+8
y-value is 35?

anonymous · Answer

6/-2=3
I don't know what you did here.

(3)^2+6*(3)+8
You are missing the negative sign out in front of x^2 here

anonymous · Answer

6/-2=3 The answer to this is -3 but still don't know why you did this

anonymous · Answer

I went and found the x coordinate of the vertex so b=6 a=-2

anonymous · Answer

because it is -x^2 there is only going to be a maximum. No minimum

anonymous · Answer

http://www.wolframalpha.com/input/?i=-x^2%2B6x%2B8&lk=4

anonymous · Answer

because you would 6/-2*1

anonymous · Answer

$$f(x)=-x^2+6x+8$$$$f'(x)=-2x+6$$$$0=-2x+6$$$$-6=-2x$$$$x=3$$$$f(3)=-(3)^2+6(3)+8$$$$f(3)=17$$$$\max=(3,17)$$

anonymous · Answer

Are you allowed to use Calculus? it is so much easier