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# 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?

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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

Oct 30, 2017

#1
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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   Oct 30, 2017
edited by CPhill  Oct 30, 2017
#3
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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

WhichWitchIsWhich  Oct 30, 2017
#2
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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   Oct 30, 2017
#4
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Sorry.....fixed it.....   Oct 30, 2017
#5
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Thanks :D 👍

WhichWitchIsWhich  Oct 30, 2017