Question

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)?

Answer 1

The equation is `y=3x^2-2x+7`

Explanation:

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

Let `f(x)=ax^2+bx+c`

`f'(x)=2ax+b`

`f'(1)=2a+b=4`, this is equation `1`

and

`f'(-1)=-2a+b=-8`, this is equation `2`

Adding the 2 equations, we get

`2b=-4`, `=>`, `b=-2`

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

`a=3`

Therefore,

`f(x)=3x^2-2x+c`

The parabola passes through `(2,15)`

So,

`f(2)=3*4-2*2+c=8+c=15`

`c=15-8=7`

Finally

`f(x)=3x^2-2x+7`

Answer 2

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

Explanation:

Given -

Point passing through `(2, 15)`

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

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

Let the equation of the parabola be -

`y=ax^2+bx+c`

We have to find the values of the parameters `a, b and c` to fix the equation.

Its slope `(dy/dx)` of the function `y=ax^2+bx+c` is defined by its first derivative.

`dy/dx=2ax+b`

Then, at `x=1;` slope `(dy/dx)=4`

Plug in these values

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

Then, at `x=-1;` slope `(dy/dx)=-8`
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 `b=-2` in equation (1)

`2a-2=4` -----------------(1)
`2a=4+2=6`
`a=6/2=3`
`a=3`

Now substitute `a=3` and `b=-2` in the equation `y=ax^2+bx+c`

`y=3x^2-2x+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(2)^2-2(2)+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=ax^2+bx+c`

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

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