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

Question

Evelyn Agard · Accepted Answer

#y = 4x - 6# First, find the derivative of #y#, which is #y = 3x^2 - 4x# Now, plug in the #x#-value (#2#) into this equation to find the slope of the tangent line: #y = 3(2)^2 - 4(2) = 4 # Now plug the #x#-value (2) back into the original equation to find the #y#-coordinate of the tangential point: #y = (2)^3 - 2(2)^2 + 2 = 2# (Technically I didn't need to do this since the #y#-coordinate was given; I did it anyway so you could find the tangent line with just a given #x#-coordinate.) Lastly, use the point-slope formula to find the equation of the line tangent to the function at #x=2#: #y-y_1 = m(x-x_1)# #y - 2= 4(x-2)# #y = 4x - 6#

Ezekiel Alexander · 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 - 2x^2 + 2 can be found by taking the derivative of each term separately.

The derivative of x^3 is 3x^2, the derivative of -2x^2 is -4x, and the derivative of 2 is 0.

So, the derivative of the function is dy/dx = 3x^2 - 4x.

To find the slope of the tangent line at the point (2,2), substitute x = 2 into the derivative:

dy/dx = 3(2)^2 - 4(2) = 12 - 8 = 4.

Therefore, the slope of the tangent line is 4.

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, we can substitute the values:

y - 2 = 4(x - 2).

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

y = 4x - 6.

Emily Ammerman · Answer

undefined To find the equation of a line tangent to the function $y = x^3 - 2x^2 + 2$ at the point $(2, 2)$, you first need to find the derivative of the function to determine its slope at that point. Then, using the point-slope form of a line equation, you can construct the equation of the tangent line. So:

1. Find the derivative of the function $y = x^3 - 2x^2 + 2$.
2. Evaluate the derivative at the point $(2, 2)$ to find the slope of the tangent line.
3. Use the point-slope form of a line equation, $y - y_1 = m(x - x_1)$, where $m$ is the slope and $(x_1, y_1)$ is the given point, to construct the equation of the tangent line.

Once you have found the slope of the tangent line and the given point, plug these values into the point-slope form equation to obtain the equation of the tangent line.