Let f(x)= -3x^4+79x^2-3x+1/ 5x^4+19x^3+2x+5. Discuss the short run behavior for f(x) and the long run behavior for f(x).

The way you type these questions without brackets, it is difficult to determine what the denominator is, as other tutors have also noted.

I will assume that 5x^4+19x^3+2x+5 is your denominator.

To check for vertical asymptotes, I set this equal to zero.
First of all I could not factor it, so I went to a reliable "equation solver".

http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=equations&s2=solve&s3=basic

gave me two complex and two real solutions.
so there are two veritical asymptotes, one around x = -3.8, another around x = -.62
when x=0 your function has value 1/5, so the y-intercept is 1/5

setting the numerator equal to zero in the same program gave me 2 real and 2 complex solutions
the reals at x=-5.15 and x=5.11
so the graph crosses the x-axis at those values
as x approaches ±infinity, your function approaches y=-3/5

(Are you supposed solve these with the use of a programmable calculator?
They seem rather unreasonable to work with otherwise)

To discuss the short run behavior and long run behavior of the function f(x) = (-3x^4 + 79x^2 - 3x + 1) / (5x^4 + 19x^3 + 2x + 5), we need to analyze the behavior of the function as x approaches positive infinity and negative infinity.

Let's begin by considering the short run behavior, which refers to the behavior of the function in the immediate vicinity of a particular point.

To analyze the short run behavior, we can examine the behavior of the function as x approaches a specific value, such as x = 0. In this case, we substitute x = 0 into the function:

f(0) = (-3(0)^4 + 79(0)^2 - 3(0) + 1) / (5(0)^4 + 19(0)^3 + 2(0) + 5) = 1 / 5

Therefore, at x = 0, the function value is 1/5.

Now, let's proceed to discuss the long run behavior, which refers to how the function behaves as x approaches positive infinity (x → ∞) or negative infinity (x → -∞).

To determine the long run behavior, we can examine the highest power of x in both the numerator and denominator of the function. In this case, the highest power of x is x^4.

For large values of x, x^4 dominates other terms in the function. Considering the highest power term, we can simplify the function as:

f(x) = (-3x^4/5x^4) + (79x^2/5x^4) + (-3x/5x^4) + (1/5x^4) / [(5x^4/5x^4) + (19x^3/5x^4) + (2x/5x^4) + (5/5x^4)]

simplifying further:

f(x) ≈ (-3/5) + (79/5x^2) + (-3/5x^3) + (1/5x^4) / (1 + (19/5x) + (2/5x^4) + (5/5x^4))

Now, as x approaches positive infinity or negative infinity, the terms with higher powers of x (x^2, x^3, x^4) become dominant. Consequently, the terms (79/5x^2), (-3/5x^3), and (1/5x^4) approach zero.

So, in the long run, as x approaches positive infinity or negative infinity, the function f(x) approaches:

f(x) ≈ (-3/5) / (1) = -3/5

Hence, in the long run, the function f(x) approaches the value of -3/5.

To summarize:
- The short run behavior of f(x) at x = 0 is f(0) = 1/5.
- The long run behavior of f(x) as x approaches positive infinity or negative infinity is f(x) ≈ -3/5.