Please help me with math

Question

Let f be the function defined by f(x) = (cx-5x^2)/(2x^2+ax+b) where a, b, and c are constants. The graph has a vertical asymptote at x=1, and f has a removable discontinuity at x = -2.

[Unless otherwise specified, the domain of a function f is assumed to be the set of all real numbers x for which f(x) is a real number.]

(a) show that a=2 and b= -4

(b) find the value of c. justify your answer.

(c) to make f continuous at x = -2, f(-2) should be defined as what value? Justify your answer.

(d) Write an equation for the horizontal asymptote to the graph of f. Show the work that leads to your answer.

*any help is appreciated.

Answer 1

The fact that there is a vertical asymptote at x = 1 means that when x = 1, the denominator is zero:

2(1)² +a(1) + b = 0

a + b = -2.

Since there is a removable discontinuity when x = -2, the binomial factor (x + 2) appears in both the numerator and the denominator when they are written in factored form. For starters f(x) can be written like this:

f(x) = x (-5x + c) / (2x² + ax + b)

Focusing on the numerator realize that the factor (-5x + c) must be -5(x+2).

-5(x + c/-5) implies that c/- 5 = 2 or c = -10.

You could also get to that result by using synthetic division by -2.

-2 | -5 c

10

----------------------

-5 c + 10

c + 10 = 0

c = - 10

Now we have:

f(x) = x(-5x - 10) / (2x² + ax +b) or -5x(x+2) / (2x² + ax + b).

To determine the other factor of the denominator besides ( x+ 2) try synthetic division by - 2. The remainder must equal zero.

-2 | 2 a b

-4 -2(a - 4)

---------------------------------

2 (a - 4) -2a + 8 + b

That means:

-2a + b = -8 combined with the earlier assertion that a + b = -2:

a + b = -2

------------------

-3a = -6

a = 2

b = -4

Now we have:

f(x) = -5x (x + 2) / (2x² + 2x - 4)

= -5x (x + 2) / 2 (x² + x - 2)

= -5x (x + 2) / 2 (x + 2) (x - 1)

After cancelling the common factor (x + 2), the residual function looks like:

g(x) = -5x / 2(x - 1)

g(-2) = 10/ -6 = -5/3, so to remove the discontinuity set f(-2) = -5/3.

Visually you can see that the horizontal asymptote is y = -5/2, because the degree of the numerator equals the degree of the denominator so the coefficients of the highest degree terms give the y value for the horizontal asymptote. Using g(x) to show your work:

g(x) = -5x/(2x - 2)

Divide every term in numerator and denominator by x:

g(x) = -5x/x / (2x/x - 2/x) = -5 / (2 - 2/x)

As x approaches ±∞ the term 2/x in the denominator approaches zero.

So as x -> ±∞, g(x) -> -5 / ( 2 - 0) = -5/2. That is, the horizontal asymptote is y = -5/2.

Visit the following graph to see all of the above:

desmos.com/calculator/egyqmcr7st