f(x)=−8x 2 −8x+96
You want the derivative?
d/dx (x^n) = n x^(n-1)
f(x)=−8x^2 −8x+96 I assume you mean
f'(x) = -16 x - 8
After reading this questions, I hopped on a tick like a hick on a moderately warm summer day
The given equation is \(f(x) = -8x^2 - 8x + 96\). It represents a quadratic function, as it contains a squared term (\(x^2\)).
To find the vertex of this quadratic function, you can use a formula called the vertex formula. The vertex formula states that for a quadratic equation in the form \(f(x) = ax^2 + bx + c\), the x-coordinate of the vertex can be found using the formula \(x = -\frac{b}{2a}\).
In this case, \(a = -8\) and \(b = -8\), so we can substitute these values into the formula to find the x-coordinate of the vertex:
\(x = -\frac{(-8)}{2(-8)} = \frac{1}{2}\)
Now that we have the x-coordinate of the vertex, we can substitute this back into the original equation to find the y-coordinate of the vertex. Plugging \(x = \frac{1}{2}\) into the equation:
\(f(\frac{1}{2}) = -8(\frac{1}{2})^2 - 8(\frac{1}{2}) + 96\)
Simplifying, we get:
\(f(\frac{1}{2}) = -8(\frac{1}{4}) - 4 + 96\)
\(f(\frac{1}{2}) = -\frac{2}{1} - 4 + 96\)
\(f(\frac{1}{2}) = -2 - 4 + 96\)
\(f(\frac{1}{2}) = 90\)
Therefore, the vertex of the quadratic function \(f(x) = -8x^2 - 8x + 96\) is \((\frac{1}{2}, 90)\).