SOLUTION: 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?

Algebra ->  Functions -> SOLUTION: 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?      Log On


   

Question 1027845: 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?
Answer by ikleyn(52795) About Me  (Show Source):
You can put this solution on YOUR website!
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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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This value of "a" is exactly the y-coordinate of the vertex of the parabola 
y = x%5E2+%2B+4x++-+31.

To find it, complete the square:

x%5E2+%2B+4x++-+31 = %28x%2B2%29%5E2+-+4+-+31 = %28x%2B2%29%5E2+-+35.

So, this value of "a" is a = -35.


 
  

   




Figure. Plot y = x%5E2+%2B+4x++-+31.