Question 619119
  <pre><font face = "Tohoma" size = 3 color = "indigo"><b> 
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(-3)}}}
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
{{{drawing(300,300,   -6, 6, -6, 10,   grid(1),
graph( 300, 300, -6, 6, -6, 10,0,6x, (x+2)^2-3 ))}}}
{{{drawing(300,300,   -6, 6, -6, 6,   grid(1),
graph( 300, 300, -6, 6, -6, 6,0, -x^2, (x+2)^2-2))}}}