Write an equation for the function shown in the graph.

If $a{x}^2+bx+c=0$, with $a\ne 0$, then $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$. Solving using factoring can be done by splitting the middle term of the equation and then making groups and equating them to zero.

From the graph, three points on the graph detected are: $(0,-3),(-1,-1),(1,-1)$.

Substitute the points $(0,-3),(-1,-1),(1,-1)$on the equation $y=a{x}^2+bx+c$to obtain three equations in variables $a,b$ and c.

1. $(x,y)=\left(0,-3\right)$

$$\begin{gathered} y=a{x}^2+bx+c \\ -3=a{\left(0\right)}^2+b\left(0\right)+c \\ c=-3 \end{gathered}$$

2. $(x,y)=\left(-1,-1\right)$

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

3. $(x,y)=\left(1,-1\right)$

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

Add equations (2) and (3) to eliminate variable b and solve for a.

$a+b+\left(a-b\right)=2+2$

$$\begin{gathered} 2a=4 \\ a=\frac{4}{2} \\ =2 \end{gathered}$$

Substitute 2 for an into equation (3) and solve for $b$.

$$\begin{gathered} a+b=2 \\ 2+b=2 \\ b=0 \end{gathered}$$

Since, $a=2;b=0;c=-3$, therefore, the required equation is $y=2x^2-3$.