Solved on Feb 24, 2024

Find the vertical asymptotes and holes of the rational function $f(x) = \frac{x-7}{x^{2}-9 x+14}$ . B. Vertical asymptote(s) at $x=3, 5$ and hole(s) at $x=7$ .

STEP 1

Assumptions
1. The given function is $f(x)=\frac{x-7}{x^{2}-9x+14}$ .
2. Vertical asymptotes occur where the denominator of a rational function is zero, and the numerator is not zero at those points.
3. Holes occur where both the numerator and the denominator are zero, but the common factor causing the zero can be canceled.

STEP 2

First, we need to factor the denominator to find the potential points where vertical asymptotes or holes may occur.
$x^{2}-9x+14 = (x-a)(x-b)$

STEP 3

To factor the quadratic expression, we look for two numbers that multiply to 14 and add up to -9. These numbers are -7 and -2.
$x^{2}-9x+14 = (x-7)(x-2)$

STEP 4

Now, we examine the numerator and the denominator to identify common factors. If there is a common factor, it indicates a hole; if there is no common factor, the zero of the denominator indicates a vertical asymptote.
$f(x)=\frac{x-7}{(x-7)(x-2)}$

STEP 5

We see that $x-7$ is a common factor in both the numerator and the denominator, which suggests there is a hole at $x=7$ .

STEP 6

Since $x-7$ is a common factor, it can be canceled out, leaving us with the simplified function:
$f(x)=\frac{1}{x-2}$

STEP 7

Now, we look for the values of $x$ that make the denominator of the simplified function zero, as these will be the vertical asymptotes.
$x-2 = 0$

STEP 8

Solve for $x$ to find the vertical asymptote.
$x = 2$

STEP 9

We have found that there is a hole at $x=7$ and a vertical asymptote at $x=2$ .
Therefore, the correct choice is B. Vertical asymptote(s) at $x=2$ and hole(s) at $x=7$ .

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Find the vertical asymptotes and holes of the rational function $f(x) = \frac{x-7}{x^{2}-9 x+14}$ . B. Vertical asymptote(s) at $x=3, 5$ and hole(s) at $x=7$ .

STEP 1

STEP 2

First, we need to factor the denominator to find the potential points where vertical asymptotes or holes may occur.
$x^{2}-9x+14 = (x-a)(x-b)$

STEP 3

To factor the quadratic expression, we look for two numbers that multiply to 14 and add up to -9. These numbers are -7 and -2.
$x^{2}-9x+14 = (x-7)(x-2)$

STEP 4

Now, we examine the numerator and the denominator to identify common factors. If there is a common factor, it indicates a hole; if there is no common factor, the zero of the denominator indicates a vertical asymptote.
$f(x)=\frac{x-7}{(x-7)(x-2)}$

STEP 5

We see that $x-7$ is a common factor in both the numerator and the denominator, which suggests there is a hole at $x=7$ .

STEP 6

Since $x-7$ is a common factor, it can be canceled out, leaving us with the simplified function:
$f(x)=\frac{1}{x-2}$

STEP 7

Now, we look for the values of $x$ that make the denominator of the simplified function zero, as these will be the vertical asymptotes.
$x-2 = 0$

STEP 8

Solve for $x$ to find the vertical asymptote.
$x = 2$

STEP 9

We have found that there is a hole at $x=7$ and a vertical asymptote at $x=2$ .
Therefore, the correct choice is B. Vertical asymptote(s) at $x=2$ and hole(s) at $x=7$ .

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Find the vertical asymptotes and holes of the rational function f(x)=x−7x2−9x+14f(x) = \frac{x-7}{x^{2}-9 x+14}f(x)=x2−9x+14x−7​. B. Vertical asymptote(s) at x=3,5x=3, 5x=3,5 and hole(s) at x=7x=7x=7.

STEP 1

STEP 2

First, we need to factor the denominator to find the potential points where vertical asymptotes or holes may occur.x2−9x+14=(x−a)(x−b) x^{2}-9x+14 = (x-a)(x-b) x2−9x+14=(x−a)(x−b)

STEP 3

To factor the quadratic expression, we look for two numbers that multiply to 14 and add up to -9. These numbers are -7 and -2.x2−9x+14=(x−7)(x−2) x^{2}-9x+14 = (x-7)(x-2) x2−9x+14=(x−7)(x−2)

STEP 4

Now, we examine the numerator and the denominator to identify common factors. If there is a common factor, it indicates a hole; if there is no common factor, the zero of the denominator indicates a vertical asymptote.f(x)=x−7(x−7)(x−2) f(x)=\frac{x-7}{(x-7)(x-2)} f(x)=(x−7)(x−2)x−7​

STEP 5

We see that x−7 x-7 x−7 is a common factor in both the numerator and the denominator, which suggests there is a hole at x=7 x=7 x=7.

STEP 6

Since x−7 x-7 x−7 is a common factor, it can be canceled out, leaving us with the simplified function:f(x)=1x−2 f(x)=\frac{1}{x-2} f(x)=x−21​

STEP 7

Now, we look for the values of x x x that make the denominator of the simplified function zero, as these will be the vertical asymptotes.x−2=0 x-2 = 0 x−2=0

STEP 8

Solve for x x x to find the vertical asymptote.x=2 x = 2 x=2

STEP 9

We have found that there is a hole at x=7 x=7 x=7 and a vertical asymptote at x=2 x=2 x=2.Therefore, the correct choice is B. Vertical asymptote(s) at x=2 x=2 x=2 and hole(s) at x=7 x=7 x=7.

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Find the vertical asymptotes and holes of the rational function $f(x) = \frac{x-7}{x^{2}-9 x+14}$ . B. Vertical asymptote(s) at $x=3, 5$ and hole(s) at $x=7$ .

First, we need to factor the denominator to find the potential points where vertical asymptotes or holes may occur.
$x^{2}-9x+14 = (x-a)(x-b)$

To factor the quadratic expression, we look for two numbers that multiply to 14 and add up to -9. These numbers are -7 and -2.
$x^{2}-9x+14 = (x-7)(x-2)$

Now, we examine the numerator and the denominator to identify common factors. If there is a common factor, it indicates a hole; if there is no common factor, the zero of the denominator indicates a vertical asymptote.
$f(x)=\frac{x-7}{(x-7)(x-2)}$

We see that $x-7$ is a common factor in both the numerator and the denominator, which suggests there is a hole at $x=7$ .

Since $x-7$ is a common factor, it can be canceled out, leaving us with the simplified function:
$f(x)=\frac{1}{x-2}$

Now, we look for the values of $x$ that make the denominator of the simplified function zero, as these will be the vertical asymptotes.
$x-2 = 0$

Solve for $x$ to find the vertical asymptote.
$x = 2$

We have found that there is a hole at $x=7$ and a vertical asymptote at $x=2$ .
Therefore, the correct choice is B. Vertical asymptote(s) at $x=2$ and hole(s) at $x=7$ .