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 ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> 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 1096609: 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(52779) About Me  (Show Source):
You can put this solution on YOUR website!
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HINT: this value of x and this value of "a" form the vertex of the given parabola. 

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


    So, this value of "a" is  a = -35,  and this value of "x" is  x = -2.