Determine the values of a, b, and c for f(x)=ax^2+bx+c so that f'(x)=6x-3 and f(2)=1

Question

Determine the values of a, b, and c for f(x)=ax^2+bx+c so that f'(x)=6x-3 and f(2)=1

Answer 1

f(x) = integral [ 6x-3] = 3x^2 - 3x + c

f(2) = 1 --> 1 = 3*2*2 - 3*2 + c

1 = 12 - 6 + c

1 = 6+c

c = -5

f(x) = 3x^2 - 3x - 5 ---> a=3 b = -3 c = -5

Answer 2

Hi Rahman W

a = 3

b = -3

c = -5

To find the coefficients in f(x) you could take the Ant-iderivative of f'(x) then to solve for the constant you plug f(2) =1 into your resulting function

f(x) = ax² + bx + c

When

f'(x) = 6x - 3

You have two exponents or powers

6x = 6(x¹)

-3 = -3(x⁰) since x⁰ = 1 you basically have -3(1)

Find the anti-derivative for f'(x) above where you have two powers n=1 and n=0

You can use the Power Rule and you can look this up in your text or online

For Powers the rule is

∫xⁿdx = xⁿ⁺¹/n +1 + c

6∫xⁿdx - 3∫xⁿdx = 6(x⁽¹⁺¹⁾)/1+1 - 3(x⁽⁰⁺¹⁾)/1 + c

6∫xⁿdx - 3∫xⁿdx = 6(x²/2) - 3(x¹/1) + c

6∫xⁿdx - 3∫xⁿdx = 3x² - 3x + c

f(x) = 3x² - 3x + c

So a = 3 and b = -3 now we to plug in f(2) to solve for c

Since f(2) = 1

1 = 3(2²) - 3(2) + c

1 = 3(4) - 3(2) + c

1 = 12 - 6 + c

1 = 6 + c

1 - 6 = c

-5 = c

so c = -5

And finally our functions

f(x) = 3x² - 3x - 5

You can graph your function at Desmos.com to confirm that f(2) is 1

I hope this helps