Question

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

Answer 1

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`

Explanation:

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`