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

$f (x) = \frac{x^{2} + 4 x}{x^{2} + 3 x - 4}$

Factor $x$ out of $x^{2} + 4 x$ .

Factor $x$ out of $x^{2}$ .

$f (x) = \frac{x \cdot x + 4 x}{x^{2} + 3 x - 4}$

Factor $x$ out of $4 x$ .

$f (x) = \frac{x \cdot x + x \cdot 4}{x^{2} + 3 x - 4}$

Factor $x$ out of $x \cdot x + x \cdot 4$ .

$f (x) = \frac{x (x + 4)}{x^{2} + 3 x - 4}$

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.

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

Cancel the common factor of $x + 4$ .

Cancel the common factor.

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

Rewrite the expression.

$f (x) = \frac{x}{x - 1}$

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

$x + 4$

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

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

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

Substitute $- 4$ for $x$ to find the $y$ coordinate of the hole.

$\frac{- 4}{- 4 - 1}$

Simplify.

Subtract $1$ from $- 4$ .

$\frac{- 4}{- 5}$

Dividing two negative values results in a positive value.

$\frac{4}{5}$

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

$(- 4, \frac{4}{5})$

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

Related Questions

Support LCMGCF.com by disabling your adblocker.