Determine the equation and vertical intercept of the line in shown the graph. The other curve is \(y=\dfrac{{3}}{{32}}x^3+\dfrac{{5}}{{16}}x^2\) and the line intersects the curve at \((2,2)\) and \((-4, y)\).
Solution: First, let's solve the mystery of what that \(y\) value is by plugging \(-4\) into the curve equation provided:
\[ \solve{ y&=&\dfrac{{3}}{{32}}(-4)^3+\dfrac{{5}}{{16}}(-4)^2\\ y&=&\dfrac{{3}}{{32}}(-64)+\dfrac{{5}}{{16}}(16)\\ y&=&3(-2)+5\\ y&=&-6+5\\ y&=&-1 } \]Now that we know the point of intersection is \( (-4,-1)\), we can calculate the slope using the other point \( (2,2) \):
\[ \solve{ m&=&\dfrac{2-(-1)}{2-(-4)}\\ m&=&\dfrac{2+1}{2+4}\\ m&=&\dfrac{{3}}{{6}}\\ m&=&\dfrac{{1}}{{2}}\\ } \]Then, we can use the Point-Slope form of the line and simplifiy to the Slope-Intercept form so we can both determine the equation of the line and the vertical intercept at the same time:
\[ \solve{ y&=&\dfrac{{1}}{{2}}\left(x-2\right)+2\\ y&=&\dfrac{{1}}{{2}}x-1+2\\ y&=&\dfrac{{1}}{{2}}x+1 } \]Thus, the equation for the line is \(y=\frac{{1}}{{2}}x+1\) and the vertical intercept is \((0,1)\).