The area between the two graphs, \begin{align} y = -x^2 + 2x + 3 \end{align} & \begin{align} y = mx + c \end{align}
is 4.5 units^2.
They intersect twice, at (-1,0) & at an unknown point. I want to find the values for m & c that make the area between the two graphs 4.5 units^2 (Refer to image for graph).
I found that m = c through:
\begin{align} \ m & = {\frac{y_2 - y_1}{x_2 - x_1}} \\ & = {\frac{c - 0}{0 - (-1)}} \\ & = {\frac{c}{1}} \\ & = c \end{align}
When I make the unknown point of intersection A(a,b) and when i find the area under both graphs separately i get:
For the parabola: \begin{align} A = \frac{-a^3}{3} + a^2 + 3a + \frac{5}{3} \end{align}
And for the linear equation, i get either: \begin{align} A = \frac{ma^2}{2} + ma - \frac{m}{2} +1 \end{align} when using integration and: \begin{align} A = \frac{1}{2} * (a+1) * (b) \end{align} When using just the formula for the area of a triangle.
and since:
\begin{align} b = -x^2 + 2x + 3 \end{align}
The area under the linear graph is:
\begin{align} A &= \frac{1}{2} * (a+1) * (-a^2 + 2a + 3 ) \\ & = \frac{-a^3+a^2+5a+3}{2} \end{align}
So: \begin{align} 4.5 = (\frac{-a^3}{3} + a^2 + 3a + \frac{5}{3} ) - (\frac{-a^3+a^2+5a+3}{2}) \end{align} which results in a = 2
And if a = 2, then b = 3
Therefore:
\begin{align} \ m & = {\frac{y_2 - y_1}{x_2 - x_1}} \\ & = {\frac{3 - 0}{2 - (-1)}} \\ & = {\frac{3}{3}} \\ & = 1 \end{align}
So m, c = 1
Take the other point if intersection to be $A=(x,y)$. Find the area under the parabola from $(-1,0)$ to $A$. Now find the same for the straight line, this is easier because this is a triangle. Subtract that latter from the former and equate that with $4.5$. You have already eliminated one variable because $m=c$. So solve for the other one.