Find the x and y Intercepts (x^2+6x)/(3x^3-17x^2-6x)

$\frac{x^{2} + 6 x}{3 x^{3} - 17 x^{2} - 6 x}$

Step 1

Write $\frac{x^{2} + 6 x}{3 x^{3} - 17 x^{2} - 6 x}$ as an equation.

$y = \frac{x^{2} + 6 x}{3 x^{3} - 17 x^{2} - 6 x}$

Step 2

Find the x-intercepts.

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

$0 = \frac{x^{2} + 6 x}{3 x^{3} - 17 x^{2} - 6 x}$

Solve the equation.

Set the numerator equal to zero.

$x^{2} + 6 x = 0$

Solve the equation for $x$ .

Factor $x$ out of $x^{2} + 6 x$ .

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

$x \cdot x + 6 x = 0$

Factor $x$ out of $6 x$ .

$x \cdot x + x \cdot 6 = 0$

Factor $x$ out of $x \cdot x + x \cdot 6$ .

$x (x + 6) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x + 6 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x + 6$ equal to $0$ and solve for $x$ .

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

The final solution is all the values that make $x (x + 6) = 0$ true.

$x = 0, - 6$

Exclude the solutions that do not make $0 = \frac{x^{2} + 6 x}{3 x^{3} - 17 x^{2} - 6 x}$ true.

$x = - 6$

x-intercept(s) in point form.

x-intercept(s): $(- 6, 0)$

Step 3

Find the y-intercepts.

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

$y = \frac{{(0)}^{2} + 6 (0)}{3 {(0)}^{3} - 17 {(0)}^{2} - 6 \cdot 0}$

The equation has an undefined fraction.

Undefined

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

y-intercept(s): $None$

Step 4

List the intersections.

x-intercept(s): $(- 6, 0)$

y-intercept(s): $None$

Step 5

Find the x and y Intercepts (x^2+6x)/(3x^3-17x^2-6x)

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