# 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)?

##### 2 Answers

The equation is

#### Explanation:

The slope at a point is

Let

and

Adding the 2 equations, we get

Therefore,

The parabola passes through

So,

Finally

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

#### Explanation:

Given -

Point passing through

Slope at

Slope at

Let the equation of the parabola be -

#y=ax^2+bx+c#

We have to find the values of the parameters

Its slope

#dy/dx=2ax+b#

Then, at

Plug in these values

#2a(1)+b=4#

#2a+b=4# -----------------(1)

Then, at

Plug in these values

#2a(-1)+b=-8#

#-2a+b=-8# -----------------(2)

Solve the equations (1) and (2) simultaneously

#2a+b=4# -----------------(1)

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

#2b=-4#

#b=(-4)/2=-2#

#b=-2#

Substitute

#2a-2=4# -----------------(1)

#2a=4+2=6#

#a=6/2=3#

#a=3#

Now substitute

#y=3x^2-2x+c#

We have to find the value of

We know the parabola is passing through the point

#3(2)^2-2(2)+c=15#

#12-4+c=15#

#8+c=15#

#c=15-8=7#

#c=7#

Now substitute

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