How do you find the equation of the line tangent to the graph of #y=x^2# at the point x=1?

Question

Ezra Adams · Accepted Answer

Find the slope by taking the derivative; use the slope and the point #(x,y) = (x,x^2)# at #x=1# to write the equation in slope-point form; then convert to standard form.
#color(white)("XXX")2x-y = 1# Given #y=x^2# then the general slope is given by the derivative #color(white)("XXX")m=(dy)/(dx) = 2x# At #x=1# #color(white)("XXX")m=(dy)/(dx) = 2(1) = 2# and #color(white)("XXX")(hatx,haty) = (1,1^2) = (1,1)# For a slope of #m# and a point #(hatx,haty)# the slope-point form of the linear equation is #color(white)("XXX")y-haty = m(x-hatx)# In this specific case #color(white)("XXX")y-1 = 2(x-1)# or #color(white)("XXX")y = 2x-1# In standard form: #color(white)("XXX") 2x-y = 1#

William Abdalla · Answer

undefined To find the equation of the line tangent to the graph of y=x^2 at the point x=1, we can use the concept of differentiation.

First, we differentiate the function y=x^2 with respect to x to find its derivative. The derivative of y=x^2 is dy/dx = 2x.

Next, we substitute x=1 into the derivative to find the slope of the tangent line at x=1. When x=1, the slope is 2(1) = 2.

Now, we have the slope of the tangent line, which is 2, and the point of tangency, which is (1, 1) since x=1.

Using the point-slope form of a linear equation, y - y1 = m(x - x1), we can substitute the values into the equation.

Therefore, the equation of the line tangent to the graph of y=x^2 at the point x=1 is y - 1 = 2(x - 1).