How do you find the equation of a line tangent to the function #y=x^3-3x^2+2# at (3,2)?

Question

Charles Anctil · Accepted Answer

# y = 9x-25 #

We have a curve given by the equation:

# y=x^3-3x^2+2 #

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. So if we differentiate the equation we have:

# dy/dx = 3x^2-6x #

And so the gradient of the tangent at #(3.2)# is given by:

# m = [dy/dx]_(x=3) #
# \ \ = 27-18 #
# \ \ = 9 #

So, using the point/slope form #y-y_1=m(x-x_1)# the tangent equations is;

# y - 2 = 9(x-3) #
# :. y - 2 =9x-27 #
# :. y = 9x-25 #

We can verify this solution graphically:

Elijah Abram · Answer

undefined To find the equation of a line tangent to a function at a given point, you need to find the derivative of the function and evaluate it at the given point.

The derivative of the function y = x^3 - 3x^2 + 2 is y' = 3x^2 - 6x.

To find the slope of the tangent line at (3,2), substitute x = 3 into the derivative: y'(3) = 3(3)^2 - 6(3) = 27 - 18 = 9.

The slope of the tangent line is 9.

Using the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope, substitute the values: y - 2 = 9(x - 3).

Simplifying the equation gives the equation of the line tangent to the function y = x^3 - 3x^2 + 2 at (3,2): y = 9x - 25.