Find the Holes in the Graph f(x)=(5x^2)/(-3x^2-7x)

$f (x) = \frac{5 x^{2}}{- 3 x^{2} - 7 x}$

Step 1

Factor $x$ out of $- 3 x^{2} - 7 x$ .

Factor $x$ out of $- 3 x^{2}$ .

$f (x) = \frac{5 x^{2}}{x (- 3 x) - 7 x}$

Factor $x$ out of $- 7 x$ .

$f (x) = \frac{5 x^{2}}{x (- 3 x) + x \cdot - 7}$

Factor $x$ out of $x (- 3 x) + x \cdot - 7$ .

$f (x) = \frac{5 x^{2}}{x (- 3 x - 7)}$

Step 2

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

Factor $x$ out of $5 x^{2}$ .

$f (x) = \frac{x (5 x)}{x (- 3 x - 7)}$

Cancel the common factors.

Cancel the common factor.

$f (x) = \frac{x (5 x)}{x (- 3 x - 7)}$

Rewrite the expression.

$f (x) = \frac{5 x}{- 3 x - 7}$

Step 3

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

$x$

Step 4

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{5 x}{- 3 x - 7}$ .

Set $x$ equal to $0$ .

$x = 0$

Substitute $0$ for $x$ in $\frac{5 x}{- 3 x - 7}$ and simplify.

Substitute $0$ for $x$ to find the $y$ coordinate of the hole.

$\frac{5 \cdot 0}{- 3 \cdot 0 - 7}$

Simplify.

Cancel the common factor of $0$ and $- 3 \cdot 0 - 7$ .

Factor $7$ out of $5 \cdot 0$ .

$\frac{7 (5 \cdot 0)}{- 3 \cdot 0 - 7}$

Cancel the common factors.

Factor $7$ out of $- 3 \cdot 0$ .

$\frac{7 (5 \cdot 0)}{7 (- 3 \cdot 0) - 7}$

Factor $7$ out of $- 7$ .

$\frac{7 (5 \cdot 0)}{7 (- 3 \cdot 0) + 7 (- 1)}$

Factor $7$ out of $7 (- 3 \cdot 0) + 7 (- 1)$ .

$\frac{7 (5 \cdot 0)}{7 (- 3 \cdot 0 - 1)}$

Cancel the common factor.

$\frac{7 (5 \cdot 0)}{7 (- 3 \cdot 0 - 1)}$

Rewrite the expression.

$\frac{5 \cdot 0}{- 3 \cdot 0 - 1}$

Multiply $5$ by $0$ .

$\frac{0}{- 3 \cdot 0 - 1}$

Simplify the denominator.

Multiply $- 3$ by $0$ .

$\frac{0}{0 - 1}$

Subtract $1$ from $0$ .

$\frac{0}{- 1}$

Divide $0$ by $- 1$ .

$0$

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

$(0, 0)$

Step 5

Find the Holes in the Graph f(x)=(5x^2)/(-3x^2-7x)

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