+267 Let f(x) = x^2 + 4x - 31. For what value of a is there exactly one real value of x such that f(x) = a?
also
Let f(x) = 3x^2 - 4x. Find the constant k such that f(x) = f(k - x) for all real numbers x.
thanks
+129899 x^2 + 4x - 31
This parabola turns upward...the x coordinate of the vertex is given by :
-4/ 2 = -2
And the y coordinate of the vertex is given by
(-2)^2 + 4(-2) - 31 = 4 - 8 - 31 = -35 = a
+267 thanks but the problem was actually -31 not +31 can you do it again please cause I dont know what to do thanks so much
+129899 We want to find
3x^2 - 4x = 3(k - x)^2 - 4(k -x)
3x^2 - 4x = 3k^2 - 6kx +3 x^2 - 4k + 4x
8x + 3k^2 - 6kx - 4k = 0
-4 ( k - 2x) + 3k ( k - 2x) = 0
(k - 2x) (3k - 4) = 0
The first factor will not produce a constant.....so.....
3k - 4 = 0
k = 4/3
