Question 252969: For what positive value of k is the graph of y = kx^2 + 6x + k +8 tangent to the x axis? Answer by drk(1908) (Show Source):
You can put this solution on YOUR website! We want to apply the following equation:
lim(h->0)
f(x+h) = k(x+h)^2 -6k + k
F(x+h) - f(x) = 2kxh + kh^2 + 6h
(F(x+h) - f(x))/h = 2xk + kh + 6
lim(h->0)F(x+h) - f(x) = 2xk + 6.
SInce we want tangent to the x-axis, y = 0. So,
K = -6/2x = -3/x.
We want positive values of k, so x = -1, or x = -3.
If x = -1, then k = 3 and
f(x) = 3x^2 + 6x + 3
f(x) = 3(x^2 + 2x + 1)
f(x) = 3(x+1)^2.
If x = -3, then k = 1, however this doesn't meet our criteria.
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