Chapter 9, Problem 29SGR

To determine

Write an equation of the graph.

Answer to Problem 29SGR

The required equation: $y=2x^2-3$

Explanation of Solution

Given:

The graph:

  

Concept Used:

From the graph we get three points on the graph: $(0,-3),(-1,-1),(1,-1)$

The standard from of the function: $y=a{x}^2+bx+c$ where a , b and c are any numbers.

Step 1: substitute the points $(0,-3),(-1,-1),(1,-1)$ on the equation $y=a{x}^2+bx+c$ . We will get 3 separate equation with a, b and c .

Calculation:

From the graph we get three points on the graph: $(0,-3),(-1,-1),(1,-1)$

The standard from of the function: $y=a{x}^2+bx+c$ where a , b and c are any numbers.

Step 1: substitute the points $(0,-3),(-1,-1),(1,-1)$ on the equation $y=a{x}^2+bx+c$ . We will get 3 separate equation with a, b and c .

  $\begin{array}{l}y=a{x}^2+bx+c\\ \text{When}x=0\text{and}y=-3\\ y=a{x}^2+bx+c\\ $ -3=a{(0)}^2+b(0)+c$c=-3\end{array}$  $\begin{array}{l}\text{When}x=-1\text{and}y=-1\\ y=a{x}^2+bx+c\\ $ -1=a{(-1)}^2+b(-1)-3$-1+3=a-b\\ 2=a-b\end{array}$  $\begin{array}{l}\text{When}x=1\text{and}y=-1\\ y=a{x}^2+bx+c\\ $ -1=a{(1)}^2+b(1)-3$-1+3=a+b\\ 2=a+b\end{array}$

Now we get 2 equations: $a+b=2\text{and}a-b=2$

Solve for a and b:

  $\begin{array}{l}\text{Add}\text{these}\text{two}\text{equations}\\ a+b=2\\ \underrightarrow{a-b=2}\\ 2a=4\\ a=\frac{4}{2}=2\end{array}$  $\begin{array}{l}\text{Solve}\text{for}b\\ a+b=2\\ 2+b=2\\ b=2-2=0\end{array}$  $a=2;b=0;c=-3$

The required equation: $y=2x^2-3$

Thus, the required equation: $y=2x^2-3$