SOLUTION: Find the value of a for which there is exactly one real value of $x$ such that f(x) = a, where f(x) = x^2 + 4x - 31 + 4x^2

SOLUTION: Find the value of a for which there is exactly one real value of $x$ such that f(x) = a, where f(x) = x^2 + 4x - 31 + 4x^2 - 13x + 25.

Question 1209316: Find the value of a for which there is exactly one real value of $x$ such that f(x) = a, where
f(x) = x^2 + 4x - 31 + 4x^2 - 13x + 25.
Answer by ikleyn(52779) (Show Source): You can put this solution on YOUR website!
.
Find the value of a for which there is exactly one real value of $x$ such that f(x) = a,
where f(x) = x^2 + 4x - 31 + 4x^2 - 13x + 25.
~~~~~~~~~~~~~~~~~~~~~~~~

                    Step by step


(1)  Reduce the given quadratic function f(x) to the standard form 

     f(x) = px^2 + qx + r  by combining like terms.



(2)  Complete the square and write your quadratic function in vertex form

        f(x) =  + s.



(3)  The value of "s" in the last formula is your answer for "a".

With given instructions, boldly go forward from the beginning to the end.

So, the value of "a" will be y-coordinate of the vertex.

Happy calculations (!)

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