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

Question

Emilia Aguilar · Accepted Answer

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]}

Aurora Alvarado · Answer

undefined To find the maximum value of the function $ f(x) = -x^2 + 8x + 7 $, you need to determine the vertex of the parabola represented by the function. The vertex of a parabola in the form $ y = ax^2 + bx + c $ is given by the point $ \left( -\frac{b}{2a}, f\left(-\frac{b}{2a}\right) \right) $. In this case, $ a = -1 $ and $ b = 8 $. Substituting these values into the vertex formula, you get:

$$ x = -\frac{b}{2a} = -\frac{8}{2(-1)} = 4 $$

Substituting $ x = 4 $ into the function, you get:

$$ f(4) = -4^2 + 8(4) + 7 = -16 + 32 + 7 = 23 $$

Therefore, the maximum value of the function $ f(x) = -x^2 + 8x + 7 $ is $ 23 $.