Find the equation of a Normal to the curve at the given point Y=X(X-2)(X+3) at X=-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
To find the equation of the normal to the curve at a given point, we need to find the derivative of the curve at that point and then use it to calculate the equation of the line.
1. Let's start by finding the derivative of the curve Y = X(X-2)(X+3).
Taking the derivative of Y with respect to X, we get:
Y' = (X^2 - 2X)(X+3) + X(X-2)
= (X^2 - 2X)(X+3) + X^2 - 2X
= X^3 + 3X^2 - 2X^2 - 6X - 2X + 6 + X^2 - 2X
= X^3 + X^2 - 10X + 6
2. Now, we substitute X = -1 into the derivative to find the slope of the tangent line at the given point.
Y'(-1) = (-1)^3 + (-1)^2 - 10(-1) + 6
= -1 + 1 + 10 + 6
= 16
3. The slope of the normal line is the negative reciprocal of the slope of the tangent line at X=-1.
So, the slope of the normal line = -1/16.
4. We have the slope of the normal line and the point (-1, Y(-1)) = (-1, -12).
Using the point-slope form of a line, the equation of the normal line is:
y - y1 = m(x - x1),
where m is the slope and (x1, y1) is the point.
Plugging in the values, we have:
y - (-12) = (-1/16)(x - (-1))
y + 12 = (-1/16)(x + 1)
y + 12 = (-1/16)x - 1/16
y = (-1/16)x - 1/16 - 12
y = (-1/16)x - 1/16 - 192/16
y = (-1/16)x - 193/16
Therefore, the equation of the normal to the curve Y = X(X-2)(X+3) at X = -1 is y = (-1/16)x - 193/16.
To find the equation of a normal to a curve at a given point, you need to follow these steps:
1. Find the derivative of the given equation to determine the slope of the tangent line.
2. Calculate the negative reciprocal of the derivative to obtain the slope of the normal line.
3. Substitute the x-coordinate of the given point into the equation obtained in step 2 to find the y-coordinate of the point of tangency.
4. Use the point-slope form of a line (y - y1 = m(x - x1)) to write the equation of the normal line, where (x1, y1) are the coordinates of the given point and m is the slope of the normal line.
Let's go through these steps to find the equation of the normal to the curve Y = X(X-2)(X+3) at X = -1:
1. Find the derivative of the given equation:
Y = X(X-2)(X+3)
Taking the derivative using the product rule, we get:
Y' = (X-2)(X+3) + X(2X-2) + X(X-2)
= (X^2 + X - 6) + (2X^2 - 2X) + (X^2 - 2X)
= 4X^2 - 3X - 6
2. Calculate the negative reciprocal of the derivative:
The slope of the normal line will be the negative reciprocal of the derivative, so we have:
m(normal) = -1 / (4X^2 - 3X - 6)
3. Substitute X = -1 into the equation obtained in step 2 to find the y-coordinate of the point of tangency:
Y(normal) = -1 / (4(-1)^2 - 3(-1) - 6)
= -1 / (4 + 3 - 6)
= -1 / 1
= -1
Therefore, the coordinates of the point of tangency are (-1, -1).
4. Write the equation of the normal line using the point-slope form:
Using the equation y - y1 = m(x - x1) and substituting the values from the point of tangency, we have:
y - (-1) = (-1 / (4X^2 - 3X - 6))(x - (-1))
y + 1 = (-1 / (4X^2 - 3X - 6))(x + 1)
This is the equation of the normal to the curve Y = X(X-2)(X+3) at X = -1.