Math / Algebra

QuestionFind the $x$ -intercepts, the $y$ -intercept, the vertical asymptotes, and the horizontal or slant asymptot of the function. Use that information to sketch a graph. Separate answers by commas, if necessary $a(x)=\frac{x^{2}-3 x+2}{x^{2}-1}$
The function has $x$ -intercept(s) at $\square$ and $y$ -intercept at $\square$ (Enter an ordered pair or a list of ordered pairs for each answer.)
The function has vertical asymptote(s) at $x=$ $\square$ The function has a Select an answer $\hat{\approx}$ asymptote at $y=$ $\square$

Studdy Solution

STEP 1

1. The function $a(x) = \frac{x^2 - 3x + 2}{x^2 - 1}$ is a rational function.
2. We need to find the $x$ -intercepts, $y$ -intercept, vertical asymptotes, and horizontal or slant asymptotes.
3. The $x$ -intercepts occur where the numerator is zero, and the denominator is non-zero.
4. The $y$ -intercept occurs when $x = 0$ .
5. Vertical asymptotes occur where the denominator is zero, and the numerator is non-zero.
6. Horizontal or slant asymptotes are determined by the degrees of the numerator and denominator.

STEP 2

1. Find the $x$ -intercepts.
2. Find the $y$ -intercept.
3. Find the vertical asymptotes.
4. Determine the horizontal or slant asymptote.

STEP 3

To find the $x$ -intercepts, set the numerator equal to zero and solve for $x$ :
$x^2 - 3x + 2 = 0$
Factor the quadratic:
$(x - 1)(x - 2) = 0$
Thus, the $x$ -intercepts are $x = 1$ and $x = 2$ .

STEP 4

To find the $y$ -intercept, evaluate the function at $x = 0$ :
$a(0) = \frac{0^2 - 3 \cdot 0 + 2}{0^2 - 1} = \frac{2}{-1} = -2$
Thus, the $y$ -intercept is $(0, -2)$ .

STEP 5

To find the vertical asymptotes, set the denominator equal to zero and solve for $x$ :
$x^2 - 1 = 0$
Factor the expression:
$(x - 1)(x + 1) = 0$
Thus, the vertical asymptotes are at $x = 1$ and $x = -1$ .

STEP 6

To determine the horizontal or slant asymptote, compare the degrees of the numerator and the denominator. Both are degree 2, so there is a horizontal asymptote at:
$y = \frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}} = \frac{1}{1} = 1$
Thus, the horizontal asymptote is $y = 1$ .
The function has $x$ -intercepts at $(1, 0)$ and $(2, 0)$ , and $y$ -intercept at $(0, -2)$ .
The function has vertical asymptotes at $x = 1$ and $x = -1$ .
The function has a horizontal asymptote at $y = 1$ .

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Studdy Solution

STEP 1

STEP 2

1. Find the $x$ -intercepts.
2. Find the $y$ -intercept.
3. Find the vertical asymptotes.
4. Determine the horizontal or slant asymptote.

STEP 3

To find the $x$ -intercepts, set the numerator equal to zero and solve for $x$ :
$x^2 - 3x + 2 = 0$
Factor the quadratic:
$(x - 1)(x - 2) = 0$
Thus, the $x$ -intercepts are $x = 1$ and $x = 2$ .

STEP 4

To find the $y$ -intercept, evaluate the function at $x = 0$ :
$a(0) = \frac{0^2 - 3 \cdot 0 + 2}{0^2 - 1} = \frac{2}{-1} = -2$
Thus, the $y$ -intercept is $(0, -2)$ .

STEP 5

To find the vertical asymptotes, set the denominator equal to zero and solve for $x$ :
$x^2 - 1 = 0$
Factor the expression:
$(x - 1)(x + 1) = 0$
Thus, the vertical asymptotes are at $x = 1$ and $x = -1$ .

STEP 6

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Studdy Solution

STEP 1

STEP 2

1. Find the x x x-intercepts.2. Find the y y y-intercept.3. Find the vertical asymptotes.4. Determine the horizontal or slant asymptote.

STEP 3

To find the x x x-intercepts, set the numerator equal to zero and solve for x x x:x2−3x+2=0 x^2 - 3x + 2 = 0 x2−3x+2=0Factor the quadratic:(x−1)(x−2)=0 (x - 1)(x - 2) = 0 (x−1)(x−2)=0Thus, the x x x-intercepts are x=1 x = 1 x=1 and x=2 x = 2 x=2.

STEP 4

To find the y y y-intercept, evaluate the function at x=0 x = 0 x=0:a(0)=02−3⋅0+202−1=2−1=−2 a(0) = \frac{0^2 - 3 \cdot 0 + 2}{0^2 - 1} = \frac{2}{-1} = -2 a(0)=02−102−3⋅0+2​=−12​=−2Thus, the y y y-intercept is (0,−2) (0, -2) (0,−2).

STEP 5

To find the vertical asymptotes, set the denominator equal to zero and solve for x x x:x2−1=0 x^2 - 1 = 0 x2−1=0Factor the expression:(x−1)(x+1)=0 (x - 1)(x + 1) = 0 (x−1)(x+1)=0Thus, the vertical asymptotes are at x=1 x = 1 x=1 and x=−1 x = -1 x=−1.

STEP 6

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1. Find the $x$ -intercepts.
2. Find the $y$ -intercept.
3. Find the vertical asymptotes.
4. Determine the horizontal or slant asymptote.

To find the $x$ -intercepts, set the numerator equal to zero and solve for $x$ :
$x^2 - 3x + 2 = 0$
Factor the quadratic:
$(x - 1)(x - 2) = 0$
Thus, the $x$ -intercepts are $x = 1$ and $x = 2$ .

To find the $y$ -intercept, evaluate the function at $x = 0$ :
$a(0) = \frac{0^2 - 3 \cdot 0 + 2}{0^2 - 1} = \frac{2}{-1} = -2$
Thus, the $y$ -intercept is $(0, -2)$ .

To find the vertical asymptotes, set the denominator equal to zero and solve for $x$ :
$x^2 - 1 = 0$
Factor the expression:
$(x - 1)(x + 1) = 0$
Thus, the vertical asymptotes are at $x = 1$ and $x = -1$ .