# Find the maximum value of the parabola f(x) = -x^2 - 2x + 9.

## Question:

Find the maximum value of the parabola {eq}f(x) = -x^2 - 2x + 9 {/eq}.

## Maximum and Minimum of the Function:

In this problem, we will find the maximum value of the given function. We will use derivative test for finding the maximum and minimum of the function.

Suppose that y = f(x) be any differentiable function.

Firstly, we will find the extremum points of the function f(x) by {eq}\displaystyle \frac{{df}}{{dx}} = 0. {/eq}

Let x = a is the extremum point of the function.

If {eq}\displaystyle {\left( {\frac{{{d^2}f}}{{d{x^2}}}} \right)_{x = a}} < 0 {/eq}, then we say that x = a is the maximum of the function f(x).

If {eq}\displaystyle {\left( {\frac{{{d^2}f}}{{d{x^2}}}} \right)_{x = a}} > 0 {/eq}, then we say that x = a is the minimum of the function f(x).

and If {eq}\displaystyle {\left( {\frac{{{d^2}f}}{{d{x^2}}}} \right)_{x = a}} = 0 {/eq}, then the function f(x) is neither maximum nor minimum. In that case the point x = a is called the saddle point of the function.

We have to find the maximum value of {eq}\displaystyle f(x) = -x^2 - 2x + 9. {/eq}

Differentiating the function f(x) with respect to x we get

{eq}\displaystyle \eqalign{ & \frac{{df}}{{dx}} = \frac{d}{{dx}}\left( { - {x^2} - 2x + 9} \right) \cr & \frac{{df}}{{dx}} = - 2x - 2 \cr & {\text{and}}\,\,\,\frac{{{d^2}f}}{{d{x^2}}} = \frac{d}{{dx}}\left( { - 2x - 2} \right) \cr & \frac{{{d^2}f}}{{d{x^2}}} = - 2. \cr} {/eq}

For extreme points, {eq}\displaystyle \frac{{df}}{{dx}} = 0 {/eq}, then

{eq}\displaystyle - 2x - 2 = 0 \Rightarrow x = - 1 {/eq}

Here, x = -1, is called the extremum point of the function f(x).

Now we will check maximum or minimum of the function f(x).

{eq}\displaystyle {\text{At}}\,\,x\, = \, - 1,\,\,{\left( {\frac{{{d^2}f}}{{d{x^2}}}} \right)_{x = - 1}} = - 2 < 0 {/eq}

Hence given function is a maximum at the point x = -1.

Now we will find the maximum value at this point.

{eq}\displaystyle \eqalign{ & f(x) = - {x^2} - 2x + 9 \cr & f( - 1) = - {( - 1)^2} - 2( - 1) + 9 \cr & f( - 1) = - 1 + 2 + 9 \cr & f( - 1) = 10. \cr} {/eq} 