g(x)=5x2-4/x+1=answer in words

verticle and horizontal asymptotes of function f(x)=3x-1/x+4

vwerticle and horizontal of g(x)=x+7/x2-4 (answer in words)

g(x)=5x2-4/x+1=answer in words

verticle and horizontal asymptotes of function f(x)=3x-1/x+4

vwerticle and horizontal of g(x)=x+7/x2-4 (answer in words)

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I am assuming this in not in a Calculus class where you have discussed derivatives. If you are familiar with derivatives, let me know and I will offer a solution using derivatives.

For a rational function a vertical asymptote occurs when a value of x makes the denominator 0 but not the numerator. Let's find that value of x.

x + 4 = 0 => x = -4

Let's see if x = -4 causes the numerator to be zero.

3*(-4) - 1 = -13 ≠ 0

So x = -4 is a vertical asymptote.

To find the horizontal asymptote, we need to determine the degree of the numerator and denominator. The degree is determined by the largest power of x in the expression. For 3x-1, the highest power is 1. It is also 1 for x+4. When the degrees are equal, then the ratio of the leading coefficients gives the horizontal asymptote. The leading coefficient is the number in front of the x with the highest power. In the numerator it is 3; in the denominator it is 1. (Remember, x = 1x not 0x.) 3/1 = 3 so the horizontal asymptote is y=3. You can verify this by graphing.

Let's look at g(x) = (x+7)/(x^2-4).

Find the values that make the denominator 0.

x^2-4=0 (I am going to solve by factoring.)

(x-2)(x+2)=0 => x-2=0 and x+2=0 => x = 2 and x = -2.

Neither 2 or -2 make the numerator 0. (Check by plugging in.) So the function as two vertical asymptotes at x = 2 and x = -2.

What is the degree of the numerator? 1

What is the degree of the denominator? 2

When the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at y=0.

Summarizing: vertical asymptotes x=2 and x=-2; horizontal asymptote y=0.

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