Question

Find a quadratic function whose x-intercepts are -1 and 5and whose range is$(-\infty ,9]$

Short answer

The quadratic equation can be given as$-x^2+4x+5$

Step-by-step solution

Given information

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

Find zeros of the function

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 $(-\infty ,9]$

That is the function approaches the negative 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

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

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

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

find the value of a

Substitute the coordinates of vertex in the equation we get

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

Substitute the values in quadratic equation and simplify it

We get

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

Conclusion

The quadratic equation can be given as

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