Expert: Paul Klarreich - 2/8/2012

Question
a) find an equation in slope-intercept form of the tangent line to the parabola y= (x-1)^2 +2 at the point (2,3)

b)1. find an equation in slope intercept form of the normal line to the parabola y= (x-1)^2 at the point (2,3)
 2. tell the coordinates of the point other than (2,3) on the parabola where this normal line intersects the parabola a second time.

c) find the equation of the tangent line and the normal line t the parabola y= (x-1)^2 +2 at its vertex

Answer
Questioner:Tommy
Country:North Carolina, United States
Category:Calculus
Private:No
Subject:calculus 1

Question:

a) find an equation in slope-intercept form of the tangent line to the parabola y= (x-1)^2 +2 at the point (2,3)

"tangent line to the"  means the slope of your line is the derivative.

So:

1. Find dy/dx.
2. Substitute  x = 2, to get m.
3. Use m, (x=2,y=3) in the point-slope form   y - y0 = m(x - x0)
4. Put it into  y = mx + b  format


b)1. find an equation in slope intercept form of the normal line to the parabola y= (x-1)^2 at the point (2,3)

"normal line to the" means "perpendicular to the tangent line", so the slope is -1/m, where you got 'm' earlier.

The rest is the same.


 2. tell the coordinates of the point other than (2,3) on the parabola where this normal line intersects the parabola a second time.

Once you get your equation y = ... , just solve simultaneous equations.  That is how you get intersections.


c) find the equation of the tangent line and the normal line t the parabola y= (x-1)^2 +2 at its vertex

The vertex is clearly at  x = 1, y = 2.  HOWEVER, if you are not asleep, you will realize that:

1. The tangent line is horizontal.
2. The normal line is vertical.

Use what you know about hor. and ver. lines.  (HINT: the answers are embedded in an earlier sentence.)