# How do you find the maximum value of y = -2x^2 + 36x - 177?

Aug 20, 2016

I got $- 15$.

Since this function is a quadratic ($a {x}^{2} + b x + c$), and since the second-degree coefficient is negative, this function has one maximum (look at the shape of any $f \left(x\right) = - | c | {x}^{2}$).

If you take the first derivative, $\frac{d}{\mathrm{dx}}$, and set the result equal to $0$, you are finding the instantaneous slope at a maximum or minimum, and you now know that it will be a maximum.

$\frac{d}{\mathrm{dx}} \left[- 2 {x}^{2} + 36 x - 177\right]$

$= - 4 x + 36$
(refer back to the Power Rule: $\frac{d}{\mathrm{dx}} \left[{x}^{n}\right] = n {x}^{n - 1}$.)

So, setting it equal to $0$:

$0 = - 4 x + 36$

$4 x = 36$

$\textcolor{g r e e n}{x = 9}$

Now that you know what $x$ value corresponds to the maximum value, the maximum value itself is the value of $f \left(x\right)$. Therefore, plug $x = 9$ into $f \left(x\right)$:

$\textcolor{b l u e}{f \left(9\right)} = - 2 {\left(9\right)}^{2} + 36 \left(9\right) - 177$

$= - 162 + 324 - 177$

$= - 339 + 324$

$= \textcolor{b l u e}{- 15}$

So, your maximum value is $f \left(9\right) = - 15$, or the coordinates of your maximum is $\textcolor{b l u e}{\left(9 \text{,} - 15\right)}$.