Limits and Derivatives, I need help with how to solve this type of question. Let f be the function defined by, f(x)=\frac{ax^2+bx+2}{2x^(2)-8}, where a and b are constants. The graph of f has a horizontal asymptote at y=3 , and f has a removable discontinuity at x=2 .

(a) Show that a=6 and b=−13 .
(b) To make f continuous at x=2 , f(2) should be defined as what value? Justify your answer.
(c) At what value of x does f have a discontinuity due to a vertical asymptote? Give a reason for your answer.

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step-by-step explanation:

step-by-step explanation:

step 1: solve one of the equations for one of the variables. let's solve the first equation for y:

step 2: substitute that equation into the other equation, and solve for x.

step 3: substitute x = 4 x = 4 x=4 into one of the original equations, and solve for y.

answer: (yeees, you need a poet.)

//blooming serenity//

beautiful in the morning

mysterious at night

wonderous nature souring

with all it's mystical might

the flower that blooms

to the fox that looms

nocturnal creatures

with various features

a blooming flower soon in sight

as a new little hatchling fears to take flight  

afraid to leave it's calming nest,

but dear little bird, mother knows best

as sure as there are stars in the sky,

life is not about who, where what, when or why-

it's the journey that follows

adventures we swallow

of   nature's deep wonders,

natural disasters and storm's thunders.

blooming serenity

life and nature, truly is heavenly.

-kerstien lehman

step-by-step explanation: