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

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

$x$-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$a>0$

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$(-1,-18)$

Substitute the coordinates of vertex in the equation we get,

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

We get,

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

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