# Thread: Finding Equation of a Parabola from Tangent Slopes

1. ## Finding Equation of a Parabola from Tangent Slopes

Find a parabola of the form given below that has slope $m_1$ at $x_1$, slope $m_2$ at $x_2$, and passes through the point $P$.

$
y = ax^2 + bx + c$

$m_1 = 9, x_1 = 10$

$m_2 = 4, x_2 = 6$

$P = (9, 2)$

2. Hello, topher0805!

Find a parabola of the form $y \:=\:ax^2+bx+c$ .that has slope $9$ at $x=10$,
slope $4$ at $x=6$, and passes through the point $(9,2)$
The slope is given by: . $y' \:=\:2ax + b$

$\begin{array}{cccccc}
x=10,\:y'=9\!: & 2a(10) + b \:=\:9 & \Rightarrow & 20a + b \:=\:9 & {\color{blue}[1]}\\
x=6,\:y' = 4\!: & 2a(6) + b \:=\:4 & \Rightarrow & 12a + b \:=\:4 & {\color{blue}[2]} \\
(9,2)\!: &&& 81a + 9b + c \:=\:2 & {\color{blue}[3]}\end{array}$

Subtract [2] from [1]: . $8a\:=\:5\quad\Rightarrow\quad\boxed{ a \:=\:\frac{5}{8}}$

Substitute into [2]: . $12\left(\frac{5}{8}\right) + b \:=\:4\quad\Rightarrow\quad\boxed{ b \:=\:-\frac{7}{2}}$

Substitute into [3]: . $81\left(\frac{5}{8}\right) + 9\left(-\frac{7}{2}\right) + c \:=\:2\quad\Rightarrow\quad\boxed{ c \:=\:-\frac{137}{8}}$

Therefore: . $\boxed{y \;=\;\frac{5}{8}x^2 - \frac{7}{2}x - \frac{137}{8}}$