Find the x and y Intercepts 1/(x^2-3)=y+1

$\frac{1}{x^{2} - 3} = y + 1$

Step 1

Find the x-intercepts.

To find the x-intercept(s), substitute in $0$ for $y$ and solve for $x$ .

$\frac{1}{x^{2} - 3} = (0) + 1$

Solve the equation.

Add $0$ and $1$ .

$\frac{1}{x^{2} - 3} = 1$

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$x^{2} - 3, 1$

Remove parentheses.

$x^{2} - 3, 1$

The LCM of one and any expression is the expression.

$x^{2} - 3$

Multiply each term in $\frac{1}{x^{2} - 3} = 1$ by $x^{2} - 3$ to eliminate the fractions.

Multiply each term in $\frac{1}{x^{2} - 3} = 1$ by $x^{2} - 3$ .

$\frac{1}{x^{2} - 3} (x^{2} - 3) = 1 (x^{2} - 3)$

Simplify the left side.

Cancel the common factor of $x^{2} - 3$ .

Cancel the common factor.

$\frac{1}{x^{2} - 3} (x^{2} - 3) = 1 (x^{2} - 3)$

Rewrite the expression.

$1 = 1 (x^{2} - 3)$

Simplify the right side.

Multiply $x^{2} - 3$ by $1$ .

$1 = x^{2} - 3$

Solve the equation.

Rewrite the equation as $x^{2} - 3 = 1$ .

$x^{2} - 3 = 1$

Move all terms not containing $x$ to the right side of the equation.

Add $3$ to both sides of the equation.

$x^{2} = 1 + 3$

Add $1$ and $3$ .

$x^{2} = 4$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{4}$

Simplify $\pm \sqrt{4}$ .

Rewrite $4$ as $2^{2}$ .

$x = \pm \sqrt{2^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 2$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 2$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 2$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 2, - 2$

x-intercept(s) in point form.

x-intercept(s): $(2, 0), (- 2, 0)$

Step 2

Find the y-intercepts.

To find the y-intercept(s), substitute in $0$ for $x$ and solve for $y$ .

$\frac{1}{{(0)}^{2} - 3} = y + 1$

Solve the equation.

Rewrite the equation as $y + 1 = \frac{1}{{(0)}^{2} - 3}$ .

$y + 1 = \frac{1}{{(0)}^{2} - 3}$

Simplify $\frac{1}{{(0)}^{2} - 3}$ .

Simplify the denominator.

Raising $0$ to any positive power yields $0$ .

$y + 1 = \frac{1}{0 - 3}$

Subtract $3$ from $0$ .

$y + 1 = \frac{1}{- 3}$

Move the negative in front of the fraction.

$y + 1 = - \frac{1}{3}$

Move all terms not containing $y$ to the right side of the equation.

Subtract $1$ from both sides of the equation.

$y = - \frac{1}{3} - 1$

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$y = - \frac{1}{3} - 1 \cdot \frac{3}{3}$

Combine $- 1$ and $\frac{3}{3}$ .

$y = - \frac{1}{3} + \frac{- 1 \cdot 3}{3}$

Combine the numerators over the common denominator.

$y = \frac{- 1 - 1 \cdot 3}{3}$

Simplify the numerator.

Multiply $- 1$ by $3$ .

$y = \frac{- 1 - 3}{3}$

Subtract $3$ from $- 1$ .

$y = \frac{- 4}{3}$

Move the negative in front of the fraction.

$y = - \frac{4}{3}$

y-intercept(s) in point form.

y-intercept(s): $(0, - \frac{4}{3})$

Step 3

List the intersections.

x-intercept(s): $(2, 0), (- 2, 0)$

y-intercept(s): $(0, - \frac{4}{3})$

Step 4

Find the x and y Intercepts 1/(x^2-3)=y+1

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