Chapter 2.1, Problem 65E

To determine

To find: The quadratic functions one that opens upward and one that opens downward.

Answer to Problem 65E

The function for the graph that open in upward direction is $f\left(x\right)=x^2-2x-3$ and the that open in downward direction is $f\left(x\right)=-x^2+2x+3$ .

Explanation of Solution

Given:

The intercepts of the graph is $\left(-1,0\right),\left(3,0\right)$ .

Calculation:

Consider the general form of the quadratic function is,

  $f\left(x\right)=a{x}^2+bx+c$ …… (1)

The parabola passes through the point $\left(-1,0\right)$ and it follows $f\left(-1\right)=0$ . So,

  $\begin{array}{l}f\left(-1\right)=a{\left(-1\right)}^2+b\left(-1\right)+c\\ 0=a-b+c\end{array}$ …… (2)

The parabola passes through the point $\left(3,0\right)$ and it follows $f\left(3\right)=0$ . So,

  $\begin{array}{l}f\left(3\right)=a{\left(3\right)}^2+b\left(3\right)+c\\ 0=9a+3b+c\end{array}$

Subtract equation (II) from the above equation.

  $\begin{array}{l}9a+3b+c-\left(a-b+c\right)=0\\ 8a+2b=0\\ a=-\frac{b}{2}\\ b=-2a\end{array}$

Substitute $-2a$ for $b$ in equation (II).

  $\begin{array}{l}a-\left(-2a\right)+c=0\\ 3a+c=0\\ a=\frac{-c}{3}\end{array}$

From above the relation between $a,b$ and $c$ is,

  $\begin{array}{l}\frac{a}{1}=-\frac{b}{2}=\frac{c}{3}\\ \frac{a}{-1}=\frac{b}{2}=\frac{c}{3}\end{array}$

Substitute the values $\frac{a}{-1}=\frac{b}{2}=\frac{c}{3}$ in equation (II). The first function for the parabola are,

  $f\left(x\right)=x^2-2x-3$

The second function for the parabola is,

  $f\left(x\right)=-x^2+2x+3$

The graph for the function $f\left(x\right)=x^2-2x-3$ is shown in Figure 1

  

Figure 1

The graph for the function $f\left(x\right)=-x^2+2x+3$ is shown in Figure 2

  

Figure 2