Chapter 2.1, Problem 55E

To determine

To find:Two quadratic functions, one open-up and other opens- down which have same given intercepts.

Answer to Problem 55E

The quadratic function having x-intercepts $\left(\text{-1,0}\right)$ and $\left(\text{3,0}\right)$ that opens up is $y=\left(x^2-2x-3\right)$ and that open down is $y=-\left(x^2-2x-3\right)$ .

Explanation of Solution

Given data: $y$

The given x-intercepts are $\left(\text{-1,0}\right)$ and $\left(\text{3,0}\right)$

Concept used:

If the graph of function $y=f\left(x\right)$ intersect x-axis at point $\left(a\text{,0}\right)$ then $x=a$ is the zero of quadratic function.

Calculations:

If $\left(a\text{,0}\right)$ and $\left(b\text{,0}\right)$ are x- intercepts of a quadratic function $y=f\left(x\right)$ , then $y=\left(x-a\right)\left(x-b\right)$ is the quadratic function that opens up, and the function $y=-\left(x-a\right)\left(x-b\right)$ opens down ..

Let $y=f\left(x\right)$ be the quadratic function that opens up and whose x- intercepts are $\left(\text{-1,0}\right)$ and $\left(\text{3,0}\right)$ . Also, $x=-1$ , and $x=3$ are the zeros of $y=f\left(x\right)$ . Therefore, the quadratic function that opens up is

Therefore, the quadratic function that opens up is

  $y=\left(x-\text{(-1)}\right)\left(x-\text{3}\right)$

  $\Rightarrow y=\left(x+1\right)\left(x-\text{3}\right)$

  $\Rightarrow y=x^2-2x-3$

The quadratic function that opens down is

  $y=-\left(x-\text{(-1)}\right)\left(x-\text{3}\right)$

  $\Rightarrow y=-\left(x^2-2x-3\right)$

We verify the results by graphing these quadratic functions.

  

Conclusion:

The quadratic function having x-intercepts $\left(\text{-1,0}\right)$ and $\left(\text{3,0}\right)$ that opens up is $y=\left(x^2-2x-3\right)$ and that open down is $y=-\left(x^2-2x-3\right)$ .