91Ó°ÊÓ

We are given the x-intercepts of a quadratic function which are$-1,5$and range is$(-∞,9]$

The quadratic function can be given as

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

As $-1,5$are the x-intercept of the function hence they are the zeros of the function On substituting we get

$f\left(x\right)=a(x+1)(x-5)$

X-intercepts of a quadratic function are the zeros of the function.

Also the range of the function is $(-∞,9]$

That is the function approaches the negative 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

$$\begin{gathered} x=\frac{-1+5}{2} \\ x=\frac{4}{2} \\ x=2 \end{gathered}$$

Also the maximum value of the function is $9$.

Hence the coordinates of vertex are$(2,9)$

Substitute the coordinates of vertex in the equation we get

$$\begin{gathered} 9=a(2+1)(2-5) \\ 9=a\left(3\right)(-3) \\ 9=-9a \\ a=-1 \end{gathered}$$

We get

$$\begin{gathered} f\left(x\right)=-(x+1)(x-5) \\ f\left(x\right)=-(x^2-4x-5) \\ f\left(x\right)=-x^2+4x+5 \end{gathered}$$

The quadratic equation can be given as

$f\left(x\right)=-x^2+4x+5$