Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 86

Find a parabola with equation $y = ax^2 + bx +c$ that has slope 4 at $x = 1$
slope -8 at $x = -1$, and passes through the point $(2,15)$


$\begin{equation} \begin{aligned} y' &= a \frac{d}{dx} (x^2) + b \frac{d}{dx}(x) + \frac{d}{dx}(c)\\ y' &= \text{slope }= 2x a + b \end{aligned} \end{equation}$


for slope 4 at $x = 1$

$\begin{equation} \begin{aligned} 4 & = 2(1) a +b\\ 4 & = 2a + b && \text{Equation 1} \end{aligned} \end{equation}$


for slope -8 at $x = -1$

$\begin{equation} \begin{aligned} -8 &= 2(-1)a + b\\ -8 &= -2a + b && \text{Equation 2} \end{aligned} \end{equation}$

We can get the other equation by substituting the given point to the given equation.

for $(2,15)$

$\begin{equation} \begin{aligned} 15 &= a(2)^2 + b(2) + c\\ 15 &= 4a + 2b + c && \text{Equation 3} \end{aligned} \end{equation}$


Combining these equations, we get

$\begin{equation} \begin{aligned} 2a + b & = 4\\ -2a + b & = - 8\\ 4a + 2b + c &= 15 \end{aligned} \end{equation}$


$\begin{equation} \begin{aligned} a &= 3\\ b &= -2\\ c &= 7 \end{aligned} \end{equation}$

Therefore, the equation of the parabola is $y = 3x^2 - 2x +7$