# what parameter is in function if f(-1)=4,f'(1)=4, definite integral between 0 and 1 of f=-1?the function is parabola

### 1 Answer | Add Yours

You need to remember the form of the equation of the quadratic:

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

Hence, you need to calculate the parameters a,b,c.

You should consider the given facts: `f(-1)=4; f'(1)=4 ; int_0^1 f(x)dx = -1.`

`` `f(-1) = a*(-1)^2 + b*(-1) + c`

`` `f(-1) = a - b + c =gt a - b + c = 4`

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

Plugging `x = 1` in `f'(x) = 2ax + b` yields:

`f'(1) = 2a + b =gt 2a + b = 4`

`int_0^1 f(x)dx = int_0^1(ax^2 + bx + c)dx = -1`

`int_0^1(ax^2 + bx + c)dx = a/3 + b/2 + c = -1`

Collecting the three equations, yields:

`a - b + c = 4`

`2a + b = 4`

`a/3 + b/2 + c = -1 =gt 2a + 3b + 6c = -6`

Adding the second equation to the first yields:

`3a + c = 8` (A)

Multiply by 3 the second equation:

`6a + 3b = 12`

Subtracting the third equation from `6a + 3b = 12` yields:

`6a + 3b - 2a - 3b - 6c= 12 + 6`

`` `4a - 6c = 18 =gt 2a - 3c = 9` (B)

Multiply the equation (A) by 3:

`9a + 3c = 24` (C)

Adding the (C) equation to (B) yields:

`9a + 3c + 2a - 3c = 24 + 9 =gt 11a = 33 =gt a = 3`

Plugging a = 3 in the second equation yields:

`6+ b = 4 =gt b = -2`

Plugging a = 3 and b = -2 in the first equation yields:

`3 + 2+ c = 4 =gt c = 4 - 5 =gt c = -1`

**The quadratic function is `f(x) = 3x^2 - 2x - 1` .**