Which graph represents the function f (x) = StartFraction 2 Over x minus 1 EndFraction + 4?
On a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = negative 4.
On a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = negative 1, and the horizontal asymptote is at y = negative 4.
On a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = negative 1, and the horizontal asymptote is at y = 4.
On a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = 4.

Answer:

On a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = 4

Step-by-step explanation:

The given function is presented as follows;

[tex]f(x) = \dfrac{2}{x - 1} + 4[/tex]

From the given function, we have;

When x = 1, the denominator of the fraction, [tex]\dfrac{2}{x - 1}[/tex], which is (x - 1) = 0, and the function becomes, [tex]\dfrac{2}{1 - 1} + 4 = \dfrac{2}{0} + 4 = \infty + 4 = \infty[/tex] therefore, the function in undefined at x = 1, and the line x = 1 is a vertical asymptote

Also we have that in the given function, as x increases, the fraction [tex]\dfrac{2}{x - 1}[/tex] tends to 0, therefore as x increases, we have;

[tex]\lim_ {x \to \infty} \dfrac{2}{(x - 1)} \to 0, and \ \dfrac{2}{(x - 1)} + 4 \to 4[/tex]

Therefore, as x increases, f(x) → 4, and 4 is a horizontal asymptote of the function, forming a curve that opens up and to the right in quadrant 1

When -∞ < x < 1, we also have that as x becomes more negative, f(x) → 4. When x = 0, [tex]\dfrac{2}{0 - 1} + 4 = 2[/tex]. When x approaches 1 from the left, f(x) tends to -∞, forming a curve that opens down and to the left

Therefore, the correct option is on a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = 4.

Answer:

On a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = 4

Step-by-step explanation:

The given function is presented as follows;

From the given function, we have;

When x = 1, the denominator of the fraction, , which is (x - 1) = 0, and the function becomes,  therefore, the function in undefined at x = 1, and the line x = 1 is a vertical asymptote

Also we have that in the given function, as x increases, the fraction  tends to 0, therefore as x increases, we have;

Therefore, as x increases, f(x) → 4, and 4 is a horizontal asymptote of the function, forming a curve that opens up and to the right in quadrant 1

When -∞ < x < 1, we also have that as x becomes more negative, f(x) → 4. When x = 0, . When x approaches 1 from the left, f(x) tends to -∞, forming a curve that opens down and to the left

Therefore, the correct option is on a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = 4.

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If x= the length of the iPad Air and y= the width of the iPad Air, the linear system of
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1
Ą
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a. (6 + 3)2 − 4
b. 23 + (14 − 4) ÷ 2

A. a. 77
b. 13
B. a. 77
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Drew is filing his tax return as single taxpayer. His taxable income is $39,000. Use the tax table provided to compute Drew's tax due and
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Single Taxpayers: Income Brackets
Tax Income
Tax Owed
Rate Bracket
1096 O to 9,525 10% of taxable income
9,526 to $952.50 plus 12% of
1296
38,700 the excess over $9,525
$4,453.50 plus 22% of
2296
38,701 to
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$38,700
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82.501 to
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157.500
$82.500
$32,089.50 plus 3296
157.501 to
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200,000
$157.500
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500,000
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Q 14. Two friends Tina and Reeta simplified two different expressions during the revision
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V2
Tina explains simplification of Torta by rationalising the denominator and Reeta
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i)
Vz
By rationalising the denominator of
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vz
a)
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b) V7(13-va)
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Vic+V5)
2
ii)
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iii)
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