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

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> 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       Log On


   

Question 1208802: 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
Answer by ikleyn(52778) About Me  (Show Source):
You can put this solution on YOUR website!
.

In this problem you have the given parabola and horizontal line y= a.


Without long conversation and discussions, they want you find the vertex of the parabola f(x) = x^2 + 4x - 31.


To find the vertex, complete the square

    x^2 + 4x - 31 = (x^2 + 4x) - 31 = (x^2 + 4x + 4) - 4 - 31 = (x+2)^2 - 35.


This is the vertex form of the given parabola equation, and it shows you that
the minimum of the parabola is -35 at the point (x,y) = (-2,-35).


The entire parabola is above the level  y = -35,  having only one point - the vertex,- with this horizontal line y= -35.


The parabola and horizontal line y = -35 have only one common point at the vertex; this point is the tangent point.


for all  b > a,  equation  f(x) = b  has two solutions;
for all  c < a,  equation  f(x) = c  has no solutions.


So, a= -35 is the ANSWER

Solved.