Hey Dayanara! Don't worry. It sounds like a lot but when you break it down, it's not so bad.
Let's start with defining some terms just to make sure we're on the same page:
Just like how a rational number is one that can be represented as a "ratio" of two integers, a rational function is a "ratio" of two polynomials. They often look like fractions. An example would be: f(x) = x2/(x-4).
An asymptote is a part of a function that infinitely approaches a certain value. If you look up the graph of the function f(x) = 1/x, you'll see a number of asymptotes. As the x values get bigger and bigger, the fraction 1/x gets closer and closer to 0, even though it will never reach it. That is a horizontal asymptote because the line stretches out horizontally. You'll also see a vertical asymptote when x=0, because the function is not defined at that point. You may recall that we are not allowed to divide by 0.
An intercept is a point where the graph of a function intersects either the x- or y-axis. You may recall that in the equation of a line, y=mx + b, b is the y-intercept because when x=0 (which describes the y-axis), the equation gives y=b. Similarly, to find the x-intercept, you would set y=0 and then solve for x.
SO! Now let's put all of that together.
We are looking for a rational function with certain vertical asymptotes (where the function is not defined) and certain x-intercepts (where y=0) and a certain y-intercept (where x=0).
If we just start with the x-intercepts, we can imagine a function like f(x) = (x-4)(x-1). When x=4, you'll get f(x) = 0*(4-1) = 0. The same goes for when x=1.
Now adding in the vertical asymptotes, we know the function is not defined at x=-5 and x=-2. Just like for f(x) = 1/x, the easiest way to do this is to make the denominator of the rational function equal to 0 at those values. That would make us have the problem of needing to divide by 0, which we are not allowed to do. A denominator like that could look like 1/[(x+5)(x+2)]. At x=-5, for example, we'd then get 1/(0*-3)=1/0=undefined.
Putting the x-intercepts and vertical asymptotes together, we get:
f(x) = [(x-4)(x-1)] / [(x+5)(x+2)]]
If you want to see what this looks like, you can plug it into a graphing calculator, like the one at desmos.com.
Finally, that y-intercept. Remember that y-intercepts are found when x=0. Plugging x=0 into the function above gives us f(x) = [(-4)(-1)] / [(5)(2)] = 4/10 = 2/5. If you used desmos to check out that equation, you'll see this matches where that graph crosses the y-axis.
So we need a way to get that equation to equal 5 when x=0 instead of 2/5. One easy way to do this is to scale the function up. To turn 2/5 into 5, you can multiply it by 25/2. The same goes for the whole function. If we multiply the whole thing by 25/2, then we will get 5 for y when x=0.
That leaves us with a final answer of:
f(x) = [25*(x-4)(x-1)] / [2*(x+5)(x+2)]]
