Answer:
[tex]a=3[/tex]
Step-by-step explanation:
The equation of the curve is
[tex]y=x^3-ax^2+bx-1[/tex].
The gradient function is given by
[tex]\frac{dy}{dx}=3x^2-2ax+b[/tex]
The gradient of this curve at the point (3,5) is equal to the gradient of the tangent [tex]y=11x-28[/tex] which is 11.
[tex]\Rightarrow 11=3(3)^2-2a(3)+b[/tex]
[tex]\Rightarrow 11=27-6a+b[/tex]
[tex]\Rightarrow b-6a=-16...(1)[/tex]
The given point, (3,5) also satisfies the equation of the curve.
[tex]\Rightarrow 3^3-a(3)^2+b(3)-1=5[/tex]
[tex]\Rightarrow 27-9a+3b-1=5[/tex]
[tex]\Rightarrow3b-9a=-21[/tex]
[tex]\Rightarrow b-3a=-7...(2)[/tex]
Equation (2) minus equation (1) gives
[tex]-3a+6a=-7+16[/tex]
[tex]3a=9[/tex]
This implies that;
[tex]a=3[/tex]
