Question

Find the equation of a Normal to the curve at the given point Y=X(X-2)(X+3) at X=-1

Answer 1

To find the equation of the normal to the curve at the given point, we need to find both the slope of the tangent line and the point of tangency.

First, let's find the slope of the tangent line. We can do this by finding the derivative of the given curve. The equation of the curve is given by Y = X(X-2)(X+3), so we differentiate with respect to X:

dY/dX = 3X^2 - 7X - 6

To find the slope of the tangent line at X = -1, we substitute X = -1 into the derivative:

m = dY/dX |_(X=-1) = 3(-1)^2 - 7(-1) - 6 = 3 + 7 - 6 = 4

So, the slope of the tangent line at X = -1 is 4.

Next, let's find the point of tangency at X = -1. We substitute X = -1 into the equation of the curve:

Y = (-1)(-1-2)(-1+3) = (-1)(-3)(2) = 6

Therefore, the point of tangency is (-1, 6).

Now, we can use the point-slope form of a line to find the equation of the normal to the curve at X = -1. The point-slope form is given by:

Y - Y1 = m(X - X1)

Substituting the values we found, we get:

Y - 6 = 4(X - (-1))

Simplifying, we have:

Y - 6 = 4(X + 1)

Expanding the brackets, we get:

Y - 6 = 4X + 4

Finally, rearranging the equation, we have the equation of the normal to the curve at X = -1:

Y = 4X + 10