How do you find the equation of the line tangent to #y=x^3 - 2x# at the point (2,4)?

Question

William Abdalla · Accepted Answer

#y=10x-16# Step 1: Take the Derivative Using the power rule and sum rule, we see that #y'=3x^2-2#. Step 2: Evaluate to find the Slope We are looking for the equation of the tangent line, and one component of that is the slope. Since the slope is the derivative at the point, we can evaluate our derivative at #x=2# to find the slope: #y'=3(2)^2-2# #y'=10# Step 3: Finding the Equation Now that we have the slope (#10#) and a point #(2,4)#, we can find the equation. Tangent lines are of the form #y=mx+b#, where #x# and #y# are points on the line, #m# is the slope, and #b# is the #y#-intercept. All we are missing is the #y#-intercept, so that's what we'll solve for: #y=mx+b# #4=10(2)+b# #4=20+b# #b=-16# Putting all the information together, the equation of the tangent line at #(2,4)# is #y=10x-16#.

Ezra Adams · Answer

undefined To find the equation of the line tangent to the curve y=x^3 - 2x at the point (2,4), we need to find the slope of the tangent line at that point.

To find the slope, we take the derivative of the function y=x^3 - 2x with respect to x.

The derivative of y=x^3 - 2x is dy/dx = 3x^2 - 2.

Substituting x=2 into the derivative, we get dy/dx = 3(2)^2 - 2 = 12 - 2 = 10.

So, the slope of the tangent line at the point (2,4) is 10.

Using the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope, we can substitute the values to find the equation of the tangent line.

Substituting (2,4) and m=10 into the equation, we get y - 4 = 10(x - 2).

Simplifying the equation, we have y - 4 = 10x - 20.

Rearranging the equation, we get y = 10x - 16.

Therefore, the equation of the line tangent to y=x^3 - 2x at the point (2,4) is y = 10x - 16.