Find the equation of the tangent of the curve y = x^3+ 4x^2-2x-3 when x = -4

To find the equation of the tangent to the curve at a given point, we need to find the slope of the tangent at that point. 

First, let’s find the derivative of the curve y = x^3 + 4x^2 – 2x – 3 with respect to x:

dy/dx = 3x^2 + 8x – 2

Next, substitute x = -4 into the derivative to find the slope of the tangent at that point:

dy/dx = 3(-4)^2 + 8(-4) – 2

       = 48 – 32 – 2

       = 14

So the slope of the tangent at x = -4 is 14.

Now, we can use the point-slope form of the equation of a line to find the equation of the tangent. The point-slope form is given by:

y – y1 = m(x – x1)

where (x1, y1) is a point on the line and m is the slope of the line.

Plugging in the point (-4, y(-4)) = (-4,5) and the slope m = 14, we get:

y – (5) = 14(x – (-4))

y -5 = 14(x + 4)

y -5 = 14x + 56

y = 14x + 56 + 5

y = 14x + 61

Therefore, the equation of the tangent to the curve y = x^3 + 4x^2 – 2x – 3 at x = -4 is y = 14x + 61.Boost your grades with an A Level Maths Revision Course