Let $f(x) = x^2 + 4x - 31$. for what value of $a$ is there exactly one real value of $x$ such that $f(x) = a$?

To solve for this, we need to find for the value of x when the 1st derivative of the equation is equal to zero (or at the extrema point).

So what we have to do first is to derive the given equation:

f (x) = x^2 + 4 x – 31

Taking the first derivative f’ (x):

f’ (x) = 2 x + 4

Setting f’ (x) = 0 and find for x:

2 x + 4 = 0

x = - 2

Therefore the value of a is:

a = f (-2)

a = (-2)^2 + 4 (-2) – 31

a = 4 – 8 – 31

a = - 35

proofs involving isosceles triangles often require special consideration because an isosceles triangle has several distinct properties that do not apply to normal triangles.(more about triangle types) therefore, when you are trying to prove that two triangles are congruent, and one or both triangles, are isosceles you have a few theorems that you can use to make your life easier.

step-by-step explanation:

an isosceles triangle has two congruent sides and two congruent angles. the congruent angles are called the base angles and the other angle is known as the vertex angle. ∠bac and ∠bca are the base angles of the triangle picture on the left. the vertex angle is ∠abc