Question

The gradient of the curve y = ax² + bx at the point (3, 3) is 4. Find the value of a and the value of b.​

Answer 1

The curve passes through the point (3, 3), so
`y=3` when
`x=3`. Then

`y = ax^2 + bx \implies 3 = 9a + 3b \implies 3a + b = 1`

The tangent line to the curve at (3, 3) has gradient
`(dy)/(dx)` at
`x=3`. Compute the derivative.

`y = ax^2 + bx \implies (dy)/(dx) = 2ax + b`

Then when
`x=3`, the gradient is 4, so

`2ax + b = 4 \implies 6a+b=4`

Solve for
`a` and
`b`. Eliminating
`b`, we find

`(6a+b) - (3a+b) = 4-1 \implies 3a = 3 \implies \boxed{a=1}`

and it follows that

`3 + b = 1 \implies \boxed{b = -2}`