ANSWER
Maximum at 7
EXPLANATION
By completing the square we will rewrite the equation in the form,
[tex]f(x)=a(x-h)^2+k[/tex]
Where (h, k) is the vertex of the parabola - in other words, it is the maximum or minimum of the function.
To complete the square we have to find a perfect square, which has the form,
[tex](a+b)^2=a^2+2ab+b^2[/tex]
In the given function a = x, so the first term is x². However, in the given function the first term is negative, so we have to take -1 as a common factor first,
[tex]f(x)=-(x^2-8x+9)[/tex]
The second term is -8x, so we can find b,
[tex]\begin{gathered} 2ab=-8x \\ if\text{ a = x} \\ 2xb=-8x\Rightarrow b=-4 \end{gathered}[/tex]
The perfect square should be,
[tex](x-4)^2=x^2-8x+16[/tex]
To have the same equation we have to subtract 16 and add 16 to the function,
[tex]f(x)=-(x^2-8x+16-16+9)=-((x-4)^2-16+9)=-((x-4)^2-7)=-(x-4)^2+7[/tex]
The vertex of the parabola is (4, 7). Hence, the maximum is at 7.
We know that it is a maximum because the coefficient a is negative, so the branches of the parabola point downward.
