Question

Find a parabola with equation y = ax^2 + bx + c that has slope 8 at x=1, slope -20 at x=-1, and passes through the point (2,17).

Answer 1

well, first off let's get the derivative of y = ax² + bx + c, keeping in mind that "a, b, c" are all scalar values or namely constants.

`y = ax^2+bx+c\implies \cfrac{dy}{dx}=2ax+b`

since the derivative of an equation is simply the slope of the tangent line at that point on the curve, then

`\begin{cases} (dy)/(dx)=8\\ x=1 \end{cases}\implies 8=2a(1)+b\implies 8-2a=b \\\\[-0.35em] ~\dotfill\\\\ \begin{cases} (dy)/(dx)=-20\\ x=-1 \end{cases}\implies -20=2a(-1)+b\implies -20=-2a+b \\\\\\ -20+2a=b\implies \stackrel{\textit{substituting from the equation above}}{-20+2a=8-2a}\implies 2a=28-2a \\\\\\ 4a=28\implies a=\cfrac{28}{4}\implies \boxed{a=7}\hspace{5em}\stackrel{ 8-2(7) }{\boxed{b=-6}} \\\\[-0.35em] ~\dotfill`

`y=7x^2-6x+c\hspace{5em}\textit{we also know that } \begin{cases} x=2\\ y=17 \end{cases} \\\\\\ 17=7(2)^2-6(2)+c\implies 17=28-12+c\implies 17=16+c\implies \boxed{1=c} \\\\\\ ~\hfill~ {\Large \begin{array}{llll} y=7x^2-6x+1 \end{array}} ~\hfill`