The curve passes through the point (3, 3), so [tex]y=3[/tex] when [tex]x=3[/tex]. Then
[tex]y = ax^2 + bx \implies 3 = 9a + 3b \implies 3a + b = 1[/tex]
The tangent line to the curve at (3, 3) has gradient [tex]\frac{dy}{dx}[/tex] at [tex]x=3[/tex]. Compute the derivative.
[tex]y = ax^2 + bx \implies \dfrac{dy}{dx} = 2ax + b[/tex]
Then when [tex]x=3[/tex], the gradient is 4, so
[tex]2ax + b = 4 \implies 6a+b=4[/tex]
Solve for [tex]a[/tex] and [tex]b[/tex]. Eliminating [tex]b[/tex], we find
[tex](6a+b) - (3a+b) = 4-1 \implies 3a = 3 \implies \boxed{a=1}[/tex]
and it follows that
[tex]3 + b = 1 \implies \boxed{b = -2}[/tex]
