What is the maximum value that the graph of $f(x)= -x^2+8x+7$ ?

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Answer 1 · 2016-07-25T18:09:30

I got $23$ .

If you think about it, since $x^2$ has a minimum, $-x^2 + bx + c$ has a maximum (that doesn't require closed bounds).

We should know that:

So, just take the derivative, set it equal to $0$ , find the value of $x$ (which corresponds to the maximum), and use $x$ to find $f(x)$ .

$f'(x) = -2x + 8$ (power rule)

$0 = -2x + 8$

$2x = 8$

$color(blue)(x = 4)$

Therefore:

$f(4) = -(4)^2 + 8(4) + 7$

$= -16 + 32 + 7$

$=> color(blue)(y = 23)$

So your maximum value is $23$ .

$y = -x^2 + 8x + 7$ :

graph{-x^2 + 8x + 7 [-7.88, 12.12, 16.52, 26.52]}

What is the maximum value that the graph of f(x)= -x^2+8x+7f(x)= -x^2+8x+7?