# Determine whether the given quadratic function has a minimum value or maximum value. Then find...

## Question:

Determine whether the given quadratic function has a minimum value or a maximum value. Then find the coordinates of the minimum or maximum point.

{eq}f(x) = x^2 + 2x - 9 {/eq}

## Maxima and Minima:

A function {eq}f(x){/eq} is said to have minima or maxima if the derivative of the function with respect to {eq}x{/eq} is zero. Now, to know whether the point is either minimum or maximum, we use the second derivative test. If the second derivative of the function is negative at a point, then the function is said to have a local maximum at that point whereas if the second derivative turns out to be positive at a point, then the function is said to have a local minimum at that point.

## Answer and Explanation:

Given Function:

$$f(x) = x^2 + 2x - 9 $$

To determine the critical point, compute the first derivative of the function and set it to zero.

$$\begin{align} \frac { d }{ dx } f\left( x \right)&=0\quad \\[0.3cm] \Rightarrow \frac { d }{ dx } \left( { x }^{ 2 }+2x-9 \right) &=0\\[0.3cm] 2x+2&=0\\[0.3cm] x&=-1 \end{align} \\ $$

This means that the given function has a critical point at {eq}x = -1 {/eq}. To determine the {eq}y-{/eq}value of the function, simply substitute {eq}x = -1 {/eq} to the original equation.

$$\begin{align} f\left( x = -1\right) &={ (-1) }^{ 2 }+2(-1)-9 \\[0.3cm] &= -10 \end{align} \\ $$

Therefore, the critical point is {eq}(-1,-10) {/eq}.

Now, perform the second derivative test to find whether the critical point is a local maxima or minima. Compute the second derivative of the function and set it to zero.

$$\begin{align} \frac { { d }^{ 2 } }{ d{ x }^{ 2 } } f\left( x \right) &=\frac { d }{ dx } \left( \frac { d }{ dx } f\left( x \right) \right) \\[0.3cm] &=\frac { d }{ dx } (2x+2) \\[0.3cm] &=2 \end{align} \\ $$

Since the second derivative is positive, the function has a minimum value at {eq}(-1,-10){/eq}.

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from General Studies Math: Help & Review

Chapter 5 / Lesson 2