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 $y = 3x^2 - 2x +7$