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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