# Thread: Find the equation of both lines?

1. ## Find the equation of both lines?

Determine the equation of both lines that are tangent to the graph of y=x^2 +4 and pass through the point (1,-2)

So would I simply use the y=mx+b form, m=2, y= -2, and x = 1?

2. Originally Posted by RyGuy
Determine the equation of both lines that are tangent to the graph of y=x^2 +4 and pass through the point (1,-2)

So would I simply use the y=mx+b form, m=2, y= -2, and x = 1?
no ... the point (1,-2) does not lie on the parabola.

3. Originally Posted by skeeter
no ... the point (1,-2) does not lie on the parabola.
Oh right.

4. Originally Posted by RyGuy
Determine the equation of both lines that are tangent to the graph of y=x^2 +4 and pass through the point (1,-2)

So would I simply use the y=mx+b form, m=2, y= -2, and x = 1?
the derivative of the parabola equation gives the slope of all tangent lines to the curve.

You need the equations of the two that pass through (1,-2).

hence, write the derivative to get the slopes of these lines.
then use

$y-y_1=m(x-x_1)$

5. point of tangency on the curve is $(x,x^2+4)
$

each tangent line passes thru the point $(1, -2)$

remember how to find the slope between two points?

... also, note that the slope of both tangent lines is y' = 2x

I'll let you take it from here.