# How do you 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 (2,15)?

Jan 2, 2017

The equation is $y = 3 {x}^{2} - 2 x + 7$

#### Explanation:

The slope at a point is $=$ the derivative.

Let $f \left(x\right) = a {x}^{2} + b x + c$

$f ' \left(x\right) = 2 a x + b$

$f ' \left(1\right) = 2 a + b = 4$, this is equation $1$

and

$f ' \left(- 1\right) = - 2 a + b = - 8$, this is equation $2$

Adding the 2 equations, we get

$2 b = - 4$, $\implies$, $b = - 2$

$2 a - 2 = 4$, from equation $1$

$a = 3$

Therefore,

$f \left(x\right) = 3 {x}^{2} - 2 x + c$

The parabola passes through $\left(2 , 15\right)$

So,

$f \left(2\right) = 3 \cdot 4 - 2 \cdot 2 + c = 8 + c = 15$

$c = 15 - 8 = 7$

Finally

$f \left(x\right) = 3 {x}^{2} - 2 x + 7$

Jan 2, 2017

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

#### Explanation:

Given -

Point passing through $\left(2 , 15\right)$

Slope at $x = 1$ is $m = 4$

Slope at $x = - 1 = - 8$ is $m = - 8$

Let the equation of the parabola be -

$y = a {x}^{2} + b x + c$

We have to find the values of the parameters $a , b \mathmr{and} c$ to fix the equation.

Its slope $\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$ of the function $y = a {x}^{2} + b x + c$ is defined by its first derivative.

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 a x + b$

Then, at x=1;  slope $\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 4$

Plug in these values

$2 a \left(1\right) + b = 4$
$2 a + b = 4$ -----------------(1)

Then, at x=-1;  slope $\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = - 8$
Plug in these values

$2 a \left(- 1\right) + b = - 8$
$- 2 a + b = - 8$ -----------------(2)

Solve the equations (1) and (2) simultaneously

$2 a + b = 4$ -----------------(1)
$- 2 a + b = - 8$ -----------------(2) Add (1) and (2)

$2 b = - 4$
$b = \frac{- 4}{2} = - 2$
$b = - 2$

Substitute $b = - 2$ in equation (1)

$2 a - 2 = 4$ -----------------(1)
$2 a = 4 + 2 = 6$
$a = \frac{6}{2} = 3$
$a = 3$

Now substitute $a = 3$ and $b = - 2$ in the equation $y = a {x}^{2} + b x + c$

$y = 3 {x}^{2} - 2 x + c$

We have to find the value of $c$

We know the parabola is passing through the point $2 , 15$. We shall use this information to find the value of $c$

$3 {\left(2\right)}^{2} - 2 \left(2\right) + c = 15$

$12 - 4 + c = 15$
$8 + c = 15$
$c = 15 - 8 = 7$
$c = 7$

Now substitute $a = 3$ , $b = - 2$ and $c = 7$in the equation $y = a {x}^{2} + b x + c$

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