Question

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 1

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.