Consider the given functions.

f (x) = - 5x^2 + 2x - 6

g (x) = - 5x^2 - 4x + 8

h (x) = 6x - 14

Select the expression that will produce h (x).

A. f (x) - g (x)

B. f (x) + f (x)

C. g (x) - f (x)

D. f (x) + g (x)

Only one answer.

neat

in h (x) we notice that there are no x^2 terms,

we also notice that in f (x), the coeffent of the x^2 term is - 5

in g (x), the coeffient of the x^2 term is also - 5

in order to eliminate the x^2 terms, we must subtract one from another (since - 5 - (-5) = 0), so the operation must be either f (x) - g (x) or g (x) - f (x)

we then can consider the constant term or the constant

if we look at the constant of h (x), we see that it is - 14

in f (x) the constant is - 6

in g (x) the constant is 8

we note that 14=-6-8, so g (x) must be multiplied by - 1 then added to f (x)

ie f (x) - g (x)

if we check the linear term

in f (x), the linear term is 2x

in g (x) the linear term is - 4x

if we do f (x) - g (x), the linear term will be 2x - (-4x) = 2x+4x=6x which checks

the operation is f (x) - g (x)

answer is A

Hello from MrBillDoesMath!

Answer: A

Discussion:

Note that h (x) does not contains an x^2 term where both f (x) and g (x) do. This means we need to subtract f (x) and g (x) to remove the x^2 term. The only choices are A and C but for C:

g (x) = - 5x^2 - 4x + 8

-f (x) = 5x^2 - 2x + 6

Adding these terms gives

(-5x^2 + 5x^2) - 4x - 2x + (8 + 6) =

0 - 6x + 14

which is NOT h (x). (It actually equals - h (x)).

The only candidate is choice A

Regards, MrB