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2 years ago

P1 = (1,5). P2 = (x,2) .p1p2=5

Mathematics

1 answer:

Alex787 [66]2 years ago

6 0

$\qquad\qquad\huge\underline{{\sf Answer}}♨$

Let's use distance formula :

$\qquad \sf \dashrightarrow \: \sqrt{(x - 1) {}^{2} + (2 - 5) {}^{2} } = 5$

$\qquad \sf \dashrightarrow \: {(x - 1) {}^{2} + ( - 3) {}^{2} } = 5 {}^{2}$

$\qquad \sf \dashrightarrow \:(x - 1) {}^{2} + 9 = 25$

$\qquad \sf \dashrightarrow \:(x - 1) {}^{2} = 16$

$\qquad \sf \dashrightarrow \:(x - 1) = 16$

$\qquad \sf \dashrightarrow \:(x - 1) = \pm4$

$\qquad \sf \dashrightarrow \:x = 5 \: \: or \: \: - 3$

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Please help 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

ivanzaharov [21]

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, and
as 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, and
as 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.