f' = 2ax
so,
9a+b = 4
-6a = 4
f(x) = -2/3 x^2 + 10
If f(x)=ax^(2)+b, f(−3)=4 and f′(−3)=4, find the coefficients a and b.
a =
b =
3 answers
from your f(x)
f'(x) = 2ax
given: f'(-3) = 4
4 = 2a(-3)
a = - 2/3
also f(-3) = 4
in original:
4 = (-2/3)(-3)^2 + b
4 + 6 = b
b = 10
check my arithmetic
f'(x) = 2ax
given: f'(-3) = 4
4 = 2a(-3)
a = - 2/3
also f(-3) = 4
in original:
4 = (-2/3)(-3)^2 + b
4 + 6 = b
b = 10
check my arithmetic
f₍x₎ = a x² + b
f '₍x₎ = 2 a x
f(- 3) = 4 means:
x = - 3 , f₍- 3₎ = 4
a x² + b = 4
a ∙ ( - 3 )² + b = 4
a ∙ 9 + b = 4
9 a + b = 4
f′(- 3) = 4 means
x = - 3 , f '₍- 3₎ = 2 a x
2 a x = 4
2 a ∙ ( - 3 ) = 4
- 6 a = 4
Divide both sides by - 6
a = - 4 / 6 =
a = 2 ∙ ( - 2 ) / 2 ∙ 3
a = - 2 / 3
Put this value in equation:
9 a + b = 4
9 ∙ ( - 2 / 3 ) + b = 4
- 18 / 3 + b = 4
- 6 + b = 4
Add 6 to both sides
b = 10
The solutions are:
a = - 2 / 3 , b = 10
f₍x₎ = - 2 / 3 x² +10
f '₍x₎ = 2 a x
f(- 3) = 4 means:
x = - 3 , f₍- 3₎ = 4
a x² + b = 4
a ∙ ( - 3 )² + b = 4
a ∙ 9 + b = 4
9 a + b = 4
f′(- 3) = 4 means
x = - 3 , f '₍- 3₎ = 2 a x
2 a x = 4
2 a ∙ ( - 3 ) = 4
- 6 a = 4
Divide both sides by - 6
a = - 4 / 6 =
a = 2 ∙ ( - 2 ) / 2 ∙ 3
a = - 2 / 3
Put this value in equation:
9 a + b = 4
9 ∙ ( - 2 / 3 ) + b = 4
- 18 / 3 + b = 4
- 6 + b = 4
Add 6 to both sides
b = 10
The solutions are:
a = - 2 / 3 , b = 10
f₍x₎ = - 2 / 3 x² +10