Question

Find a quadratic function whose x-intercepts are -4 and 2 and whose range is$[-18,\infty )$

Short answer

The quadratic function is$f\left(x\right)=2x^2+4x-16$

Step-by-step solution

Given information

The x-intercepts of the function function are$-4,2$. And range of the function is$[-18,\infty )$

The zeros of the function

$x$-intercepts of a quadratic function are the zeros of the function

The quadratic function can be given as

$f\left(x\right)=a(x-x_1)(x-x_2)$

As $-4,2$are the x-intercept of the function hence they are the zeros of the function

On substituting we get

$f\left(x\right)=a(x+4)(x-2)$

Also the range of the function is $[-18,\infty )$

That is the function approaches the positive infinity as the values of a changes From this we can conclude that$a>0$

Find the value of the vertex

Now We know that the quadratic equation is symmetric around the vertex

This means that the zeros will be equidistant from the vertex

Hence the vertex can be given by midpoint formula

$$\begin{gathered} x=\frac{-4+2}{2} \\ x=\frac{-2}{2} \\ x=-1 \end{gathered}$$

Also the minimum value of the function is $-18$

Hence the coordinates of vertex are$(-1,-18)$

Find the value of a

Substitute the coordinates of vertex in the equation we get,

$$\begin{gathered} -18=a(-1+4)(-1-2) \\ -18=a\left(3\right)(-3) \\ -18=-9a \\ a=2 \end{gathered}$$

Substitute the values and find the quadratic equation

We get,

$$\begin{gathered} f\left(x\right)=2(x+4)(x-2) \\ f\left(x\right)=2(x^2+2x-8) \\ f\left(x\right)=2x^2+4x-16 \end{gathered}$$

Conclusion

The quadratic function can be given as$f\left(x\right)=2x^2+4x-16$