Step1

Part (a):

Use the graph to write the solution to the system.

There are one quadratic equation and one linear equation.

The intersecting point of both of the graph is the solution of the function.

$(-3,7)(3,5)$

Step 2

Part (b):

The equation of the linear function is passing through the point (-3, 7) and (3, -5):

line $y=mx+b\text{}passing\text{}through(-3,7).(3.-5)$

slope: $m=\frac{-5-7}{3-(-3)}=-2$

$y=(-2)x+b$

Plug in (-3,7) in the equation $y=(-2)x+b$:

$7=(-2)(-3)+b\Rightarrow b=1$

$y=-2x+1$

Step 3

The equation of quadratic function $a{x}^{2}+bx=0$ on the grid is passing through (-2, 0), (4, 0) and (0, -8):

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

passing through (-2,0),(4,0) and (0,-8):

$0=a{(-2)}^{2}+b(-2)+c\Rightarrow 4a-2b+c=0$

$0=a\left(4\right)+b{\left(4\right)}^{2}+c\Rightarrow 16a+4b+c=0$

$-8=a\left(0\right)+b\left(0\right)+c\Rightarrow c=-8$

$$\left[\begin{array}{c}4a-2b-8=0\\ 16a+4b-8=0\end{array}\right]\Rightarrow b=-2,a=1$$

$a=1,b=-2,c=-8$

$y\left(x\right)={x}^{2}-2x-8$