The straight line with equation y = 11x - 28 is a tangent to the curve y = {x}^(3) - a {x}^(2) + bx - 1 at the point (3,5). Find A, the coefficient of {x}^(2)

Question

asked Apr 11, 2020 91.9k views

1 Answer

Kyle Robinson Young · Answer 1 · 2020-04-16T23:51:12+0000

Answer:

a=3

Explanation:

The equation of the curve is

y=x^3-ax^2+bx-1 .

The gradient function is given by

(dy)/(dx)=3x^2-2ax+b

The gradient of this curve at the point (3,5) is equal to the gradient of the tangent
y=11x-28 which is 11.

$\Rightarrow 11=3(3)^2-2a(3)+b$

$\Rightarrow 11=27-6a+b$

$\Rightarrow b-6a=-16...(1)$

The given point, (3,5) also satisfies the equation of the curve.

$\Rightarrow 3^3-a(3)^2+b(3)-1=5$

$\Rightarrow 27-9a+3b-1=5$

$\Rightarrow3b-9a=-21$

$\Rightarrow b-3a=-7...(2)$

Equation (2) minus equation (1) gives

-3a+6a=-7+16

3a=9

This implies that;

a=3