Find the Holes in the Graph y=(x^2+3x-4)/(x^2-1)

$y = \frac{x^{2} + 3 x - 4}{x^{2} - 1}$

Factor $x^{2} + 3 x - 4$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 4$ and whose sum is $3$ .

$- 1, 4$

Write the factored form using these integers.

$y = \frac{(x - 1) (x + 4)}{x^{2} - 1}$

Factor $x^{2} - 1$ .

Rewrite $1$ as $1^{2}$ .

$y = \frac{(x - 1) (x + 4)}{x^{2} - 1^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$y = \frac{(x - 1) (x + 4)}{(x + 1) (x - 1)}$

Cancel the common factor of $x - 1$ .

Cancel the common factor.

$y = \frac{(x - 1) (x + 4)}{(x + 1) (x - 1)}$

Rewrite the expression.

$y = \frac{x + 4}{x + 1}$

To find the holes in the graph, look at the denominator factors that were cancelled.

$x - 1$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{x + 4}{x + 1}$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Substitute $1$ for $x$ in $\frac{x + 4}{x + 1}$ and simplify.

Substitute $1$ for $x$ to find the $y$ coordinate of the hole.

$\frac{1 + 4}{1 + 1}$

Simplify.

Add $1$ and $4$ .

$\frac{5}{1 + 1}$

Add $1$ and $1$ .

$\frac{5}{2}$

The holes in the graph are the points where any of the cancelled factors are equal to $0$ .

$(1, \frac{5}{2})$

Find the Holes in the Graph y=(x^2+3x-4)/(x^2-1)

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