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

(1) is the equation y=x³+4x²-2x-3. First we want to find the coordinates of the curve for x=-4. This means that we are going to substitute x=-4 into (1) to find the value of y. y=(-4)³+ 4(-4)²- 2(-4) -3 , so y = 5. This means that the curve passes through the point (-4, 5) and the tangent also passes through this point. The tangent is going to be in the form y=ax+b (as the tangent is a straight line) where a is the gradient of the tangent. a is also the gradient of the curve at the point x=-4. To find a, we need to differentiate the equation for the curve , and substitute in the value of x=-4. Differentiating (1) we get dy/dx = 3x²+8x-2. Substituting in x=-4, dy/dx = 3(-4)²+8(-4)-2, so dy/dx= 14. This means that a=14. We now know the gradient of the tangent (14) and a point which the tangent passes through (-4,5). Using the formula for a point on a straight line with coordinates (c,d) and gradient m, (y-d)/(x-c) = m. So (y-5) / (x-(-4))= 14. Multiplying both sides of the equation by (x+4) gives y-5 = 14 (x+4). Expanding the brackets gives y-5 = 14x + 56. Adding 5 to both sides gives y = 14x + 61. This is in the form y = ax+b, so the answer is y = 14x+61.