Question

1. Find the equation in the form y= ax^2 + bx + c that crosses the x-axis at (-4,0) and (9,0) and also passes through the point (1, 120)

2. Hence find where the curve crosses the y-axis

Answer 1

You plug in the points into the equation and solve the system:
\[\begin{cases} \begin{align}0&=a(-4)^2+b(-4)+c\\
0&=a(9)^2+b(9)+c\\
120&=a(1)^2+b(1)+c\\
\end{align}
\end{cases}\]

Answer 2

Ok can you keep going I understand where you got to but I dont know how to do the next step

Answer 3

Once you solve the system for a, b, and c, then you can substitute x=0 to determine the y-intercept.