SOLUTION: The equation is y=(x+2)^2+k where k is a constant. a)Find k when its tangent at x=1 passes through (0,0). b)Find k when the quadratic is tangent to y=-x^2

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: The equation is y=(x+2)^2+k where k is a constant. a)Find k when its tangent at x=1 passes through (0,0). b)Find k when the quadratic is tangent to y=-x^2      Log On


   

Question 619119: The equation is y=(x+2)^2+k where k is a constant.
a)Find k when its tangent at x=1 passes through (0,0).
b)Find k when the quadratic is tangent to y=-x^2
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi, Re TY, Note: completed the exercise by demonstrating using the 1st  derivative.

The equation is y=(x+2)^2+k where k is a constant.

a)Find k when its tangent at x=1 passes through (0,0). 

y=(x+2)^2+k , y' = 2(x+2), x=1, m = 6, tangent line is y = 6x, Tangent at(1,6)

6 = (1+2)^2 + k, k = highlight%28-3%29

b)Find k when the quadratic is tangent to y= -x^2, k = -2 (see 2nd graph)

 Taking 1st derivatives of both:  2(x+2) = -2x, x = -1, tangent at (-1,-1)

 -1 = (-1+2)^2 + k, k = -2