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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
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You plug in the points into the equation and solve the system: \[\begin{cases} \begin{align}0&=a(-4)^2+b(-4)+c\\ 0&=a(9)^2+b(9)+c\\ 120&=a(1)^2+b(1)+c\\ \end{align} \end{cases}\]
Ok can you keep going I understand where you got to but I dont know how to do the next step
Once you solve the system for a, b, and c, then you can substitute x=0 to determine the y-intercept.
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