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IrinaK [193]
3 years ago
10

Give an example of a rational function that has a horizontal asymptote at y = 1 and a vertical asymptote at x = 4.

Mathematics
1 answer:
Greeley [361]3 years ago
4 0

Answer:

Possibility 1: \frac{(x+1)(x+1)}{(x-4)(x+5)}

Possibility 2: \frac{(x+1)(x+3)(x-3)}{x(x-4)(x+5)}

Possibility 3: \frac{-4(x-4)(x+1)}{-4(x-4)(x-4)}

There are infinitely many more possibilities.

Step-by-step explanation:

So we are looking for a fraction in terms of x.

We have a horizontal asymptote so that means the degree of the top has to equal to degree of the bottom. It is also at y=1 which means the coefficient of the leading term on top and bottom must be the same (but not zero) since the same number divided by the same number is 1.  

Now we also have a vertical asymptote at x=4 which means we need a factor of x-4 on bottom.

So there is a lot of possibilities. Here are a few:

Possibility 1: \frac{(x+1)(x+1)}{(x-4)(x+5)}

You have the degrees are the same on top and bottom and the leading coefficients are the same.  You also have that factor of (x-4) on bottom.

Possibility 2: \frac{(x+1)(x+3)(x-3)}{x(x-4)(x+5)}

Possibility 3: \frac{-4(x-4)(x+1)}{-4(x-4)(x-4)}

Now a factor of (x-4) can be canceled here but you still have a factor of (x-4) left on bottom so you still have the vertical asymptote. You also still have the same leading coefficient on top and bottom (-4 in this case) and the same degree.

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7. Use the quadratic formula to find the solution(s). x² + 2x - 8 = 0​
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Hey there!

<u>Use the quadratic formula to find the solution(s). x² + 2x - 8 = 0</u>

  • Answer :

x = -4 or x = 2 ✅

  • Explanation :

<em><u>Quadratic</u></em><em><u> </u></em><em><u>formula </u></em><em><u>:</u></em><em><u> </u></em>ax² + bx + c = 0 where a ≠ 0

The number of real-number solutions <em>(roots)</em> is determined by the discriminant (b² - 4ac) :

  • If b² - 4ac > 0 , There are 2 real-number solutions

  • If b² - 4ac = 0 , There is 1 real-number solution.

  • If b² - 4ac < 0 , There is no real-number solution.

The <em><u>roots</u></em> of the equation are determined by the following calculation:

x =  \frac{ - b \pm  \sqrt{ {b}^{2} - 4ac } }{2a}

Here, we have :

  • a = 1
  • b = 2
  • c = -8

1) <u>Calculate </u><u>the </u><u>discrim</u><u>i</u><u>n</u><u>ant</u><u> </u><u>:</u>

b² - 4ac ⇔ 2² - 4(1)(-8) ⇔ 4 - (-32) ⇔ 36

b² - 4ac = 36 > 0 ; The equation admits two real-number solutions

2) <u>Calculate </u><u>the </u><u>roots </u><u>of </u><u>the </u><u>equation</u><u>:</u>

▪️ (1)

x_1 =  \frac{ - b -  \sqrt{ {b}^{2}  - 4ac} }{2a}  \\  \\ x_1 =  \frac{ - 2 -  \sqrt{36} }{2(1) }  \\  \\ x_1 =  \frac{ - 2 - 6}{2}   \\ \\ x_1 =  \frac{ - 8}{2}  \\  \\ \blue{\boxed{\red{x_1 = -4}}}

▪️ (2)

x_2 =  \frac{ - b  +   \sqrt{ {b}^{2} - 4ac } }{2a}  \\  \\ x_2 =  \frac{ - 2 +  \sqrt{36} }{2(1)}  \\  \\ x_2 =  \frac{ - 2 + 6}{2}  \\  \\ x_2 =  \frac{4}{2}  \\  \\ \red{\boxed{\blue{x_2 = 2}}}

>> Therefore, your answers are x = -4 or x = 2.

Learn more about <u>quadratic equations</u>:

brainly.com/question/27638369

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