SOLUTION: the graph of the function f(x)=x^2-kx+k+8 touches the x-axis at one point. what are the possible values of k?
Thanks in advance!
Algebra.Com
Question 376362: the graph of the function f(x)=x^2-kx+k+8 touches the x-axis at one point. what are the possible values of k?
Thanks in advance!
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
Since
f(x)=x^2-kx+k+8
touches the x-axis at only one point, it must be the vertex.
.
this must mean the "determinant" is zero:
b^2-4ac = 0
(-k)^2 - 4(1)(k+8) = 0
(-k)^2 - 4(k+8) = 0
k^2 - 4k-32 = 0
(k-8)(k+4) = 0
.
k = {-4, 8}
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