# Find the equation of both lines?

• Mar 21st 2010, 04:43 PM
RyGuy
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?
• Mar 21st 2010, 04:47 PM
skeeter
Quote:

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.
• Mar 21st 2010, 04:50 PM
RyGuy
Quote:

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

Oh right.

• Mar 21st 2010, 04:57 PM
Quote:

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

\$\displaystyle y-y_1=m(x-x_1)\$
• Mar 21st 2010, 05:05 PM
skeeter
point of tangency on the curve is \$\displaystyle (x,x^2+4)
\$

each tangent line passes thru the point \$\displaystyle (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.