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

first anwser choice.

step-by-step explanation:

the greater/less or equal to symbol (> or < with the line at the bottom) is a closed/filled in circle.

-5c + 2 _< _27

subtract from both sides by 2

-5c _< _ 27

divide both sides by -5

c _< _ 5.4

flip the inequality sign

c _> _5.4

the only possible answer that would match the simplified inequality is the first answer choice

the formula of a slope:

from a table we have

substitute:

the slope of f(x) is -1.

the graph of funcfion g(x) is a horizontal line. the slope of a horizontal line is equal 0. therefore the slope of g(x) is 0.

-1 < 0