Chapter 2.1, Problem 67E

To determine

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

Answer to Problem 67E

The function for the graph that open in upward direction is $f\left(x\right)=x^2-10x$ and the that open in downward direction is $f\left(x\right)=-x^2+10x$ .

Explanation of Solution

Given:

The intercepts of the graph is $\left(0,0\right),\left(10,0\right)$ .

Calculation:

Consider the general form of the quadratic function is,

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

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

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

The parabola passes through the point $\left(10,0\right)$ and it follows $f\left(10\right)=0$ . So,

  $\begin{array}{l}f\left(10\right)=a{\left(10\right)}^2+b\left(10\right)+c\\ 0=100a+10b+c\end{array}$

Substitute $0$ for $c$ in the above equation.

  $\begin{array}{l}0=100a+10b+0\\ 100a+10b=0\\ b=-10a\end{array}$ …… (II)

Substitute $-1$ for $a$ in equation (II).

  $\begin{array}{c}b=-10\left(-1\right)\\ =10\end{array}$

Substitute $1$ for $a$ in equation (II).

  $\begin{array}{c}b=-10\left(1\right)\\ =-10\end{array}$

Substitute $0$ for $c$ , $1$ for $a$ and $-10$ for $b$ in equation (I).

  $f\left(x\right)=x^2-10x$

Substitute $0$ for $c$ , $-1$ for $a$ and $10$ for $b$ in equation (I).

  $f\left(x\right)=-x^2+10x$

The graph for the function $f\left(x\right)=x^2-10x$ is shown in Figure 1

  

Figure 1

The graph for the function $f\left(x\right)=-x^2+10x$ is shown in Figure 2

  

Figure 2