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avatar+272 

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
avatar+98005 
+1

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

 

 

cool cool cool

 Oct 30, 2017
edited by CPhill  Oct 30, 2017
 #3
avatar+272 
0

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
avatar+98005 
+1

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

 

 

cool cool cool

 Oct 30, 2017
 #4
avatar+98005 
0

Sorry.....fixed it.....

 

 

cool cool cool

 Oct 30, 2017
 #5
avatar+272 
0

Thanks :D 👍

WhichWitchIsWhich  Oct 30, 2017

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