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VARVARA [1.3K]
3 years ago
5

What are the vertical asymptotes of the function f(x) = the quantity of 4 x plus 8, all over x squared plus 3 x minus 4? x = −1

and x = −2 x = −1 and x = 2 x = −1 and x = −4 x = 1 and x = −4
Mathematics
1 answer:
asambeis [7]3 years ago
6 0

To find a rational function's vertical asymyptotes, you need to find when the denominator is zero. So when is the denominator x² + 3x - 4 zero? We set that denominator equal to zero.

x² + 3x - 4 = 0

This second degree equation factors. We want a pair of numbers that multiply to 4, and there are only two pairs: 2 and 2, 1 and 4. We want the sum to 3, and one must be negative. -2 and 2 add to zero, -4 and 1 add to -3, -1 and 4 add to 3. The last one is the pair.

x² + 3x - 4 = 0

(x + 4) (x - 1) = 0

x = -4 or x = 1

So the function's vertical asymptotes are at x = 1 and x = -4.

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Answer:

Step-by-step explanation:

Let X is the number of cards Kenny has

Let Y is the number of cards Lee has

1. Kenny had 5/7 more trading cards than lee, it means

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2. After Kenny collected 50% more trading cards, he had 220 more trading cards than lee, it means:

X + 50%X = Y +  220 (2)

So, we solve the 2 equations to find out X and Y

\left \{ {{X=Y + 5/7X} \atop {1.5X=Y +220}} \right.

<=> X = \frac{3080}{17} and Y = \frac{880}{17}

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