Mathematics, 29.06.2019 07:10, coleman310

Me! this is is rational function and i don’t know how to/ don’t remember how do this! how would i find and write the equation for it?

Me! this is is rational function and i don’t know how to/ don’t remember how do this! how would i

Answers: 1

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answered: rockyroad19

An answer is

$\displaystyle f\left(x\right)=\frac{\left(x+1\right)^3}{\left(x+2\right)^2\left(x-1\right)}$

Explanation:

Template:

$\displaystyle f(x) = a \cdot \frac{(\cdots) \cdots (\cdots)}{( \cdots )\cdots( \cdots )}$

There is a nonzero horizontal asymptote which is the line y = 1. This means two things: (1) the numerator and degree of the rational function have the same degree, and (2) the ratio of the leading coefficients for the numerator and denominator is 1.

The only x-intercept is at x = -1, and around that x-intercept it looks like a cubic graph, a transformed graph of y = x^3 ; that is, the zero looks like it has a multiplicty of 3. So we should probably put (x+1)^3 in the numerator.

We want the constant to be a = 1 because the ratio of the leading coefficients for the numerator and denominator is 1. If a was different than 1, then the horizontal asymptote would not be y = 1.

So right now, the function should look something like

$\displaystyle f(x) = \frac{(x+1)^3}{( \cdots )\cdots( \cdots )}.$

Observe that there are vertical asymptotes at x = -2 and x = 1. So we need the factors (x+2)(x-1) in the denominator. But clearly those two alone is just a degree-2 polynomial.

We want the numerator and denominator to have the same degree. Our numerator already has degree 3; we would therefore want to put an exponent of 2 on one of those factors so that the degree of the denominator is also 3.

A look at how the function behaves near the vertical asympotes gives us a clue.

Observe for x = -2,

as x approaches x = -2 from the left, the function rises up in the positive y-direction, andas x approaches x = -2 from the right, the function rises up.

Observe for x = 1,

as x approaches x = 1 from the left, the function goes down into the negative y-direction, andas x approaches x = 1 from the right, the function rises up into the positive y-direction.

We should probably put the exponent of 2 on the (x+2) factor. This should help preserve the function's sign to the left and right of x = -2 since squaring any real number always results in a positive result.

So now the function looks something like

$\displaystyle f(x) = \frac{(x+1)^3}{(x+2 )^2(x-1)}.$

If you look at the graph, we see that f(-3) = 2 . Sure enough

$\displaystyle f(-3) = \frac{(-3+1)^3}{(-3+2 )^2(-3-1)} = \frac{-8}{(1)(-4)} = 2.$

And checking the y-intercept, f(0),

$\displaystyle f(0) = \frac{(0+1)^3}{(0+2 )^2(0-1)} = \frac{1}{4(-1)} = -1/4 = -0.25.$

and checking one more point, f(2),

$\displaystyle f(2) = \frac{(2+1)^3}{(2+2 )^2(2-1)} = \frac{27}{(16)(1)} \approx 1.7$

So this function does seem to match up with the graph. You could try more test points to verify.

======

If you're extra paranoid, you can test the general sign of the graph. That is, evaluate f at one point inside each of the key intervals; it should match up with where the graph is. The intervals are divided up by the factors:

x < -2. Pick a point in here and see if the value is positive, because the graph shows f is positive for all x in this interval. We've already tested this: f(-3) = 2 is positive.-2 < x < -1. Pick a point in here and see if the value is positive, because the graph shows f is positive for all x in this interval.-1 < x < 1. Pick a point here and see if the value is negative, because the graph shows f is negative for all x in this interval. Already tested since f(0) = -0.25 is negative.x > 1. See if f is positive in this interval. Already tested since f(2) = 27/16 is positive.

So we need to see if -2 < x < -1 matches up with the graph. We can pick -1.5 as the test point, then

$\displaystyle f(-1.5) = \frac{\left(-1.5+1\right)^3}{\left(-1.5+2\right)^2\left(-1.5-1\right)} = \frac{(-0.5)^3}{(0.5)^2(-2.5)} \\= (-0.5)^3 \cdot \frac{1}{(0.5)^2} \cdot \frac{1}{-2.5}$

We don't care about the exact value, just the sign of the result.

Since (-0.5)^3 is negative, (0.5)^2 is positive, and (-2.5) is negative, we really have a negative times a positive times a negative. Doing the first two multiplications first, (-) * (+) = (-) so we are left with a negative times a negative, which is positive. Therefore, f(-1.5) is positive.

answered: Guest

im not quite sure if this is right but i got 5.6 oisutuve

step-by-step explanation:

answered: Guest

Then just do it (lol)

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Me! this is is rational function and i don’t know how to/ don’t remember how do this! how would i...