Find the x and y Intercepts (2x^2-5x-12)/(x-4)

$\frac{2 x^{2} - 5 x - 12}{x - 4}$

Step 1

Write $\frac{2 x^{2} - 5 x - 12}{x - 4}$ as an equation.

$y = \frac{2 x^{2} - 5 x - 12}{x - 4}$

Step 2

Find the x-intercepts.

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

$0 = \frac{2 x^{2} - 5 x - 12}{x - 4}$

Solve the equation.

Set the numerator equal to zero.

$2 x^{2} - 5 x - 12 = 0$

Solve the equation for $x$ .

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 12 = - 24$ and whose sum is $b = - 5$ .

Factor $- 5$ out of $- 5 x$ .

$2 x^{2} - 5 x - 12 = 0$

Rewrite $- 5$ as $3$ plus $- 8$

$2 x^{2} + (3 - 8) x - 12 = 0$

Apply the distributive property.

$2 x^{2} + 3 x - 8 x - 12 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 x^{2} + 3 x) - 8 x - 12 = 0$

Factor out the greatest common factor (GCF) from each group.

$x (2 x + 3) - 4 (2 x + 3) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 x + 3$ .

$(2 x + 3) (x - 4) = 0$

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

$2 x + 3 = 0$

$x - 4 = 0$

Set $2 x + 3$ equal to $0$ and solve for $x$ .

Set $2 x + 3$ equal to $0$ .

$2 x + 3 = 0$

Solve $2 x + 3 = 0$ for $x$ .

Subtract $3$ from both sides of the equation.

$2 x = - 3$

Divide each term in $2 x = - 3$ by $2$ and simplify.

Divide each term in $2 x = - 3$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Set $x - 4$ equal to $0$ and solve for $x$ .

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

The final solution is all the values that make $(2 x + 3) (x - 4) = 0$ true.

$x = - \frac{3}{2}, 4$

Exclude the solutions that do not make $0 = \frac{2 x^{2} - 5 x - 12}{x - 4}$ true.

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

x-intercept(s) in point form.

x-intercept(s): $(- \frac{3}{2}, 0)$

Step 3

Find the y-intercepts.

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

$y = \frac{2 {(0)}^{2} - 5 \cdot 0 - 12}{(0) - 4}$

Solve the equation.

Remove parentheses.

$y = \frac{2 \cdot 0^{2} - 5 \cdot 0 - 12}{(0) - 4}$

Remove parentheses.

$y = \frac{2 \cdot 0^{2} - 5 \cdot 0 - 12}{0 - 4}$

Remove parentheses.

$y = \frac{2 {(0)}^{2} - 5 \cdot 0 - 12}{(0) - 4}$

Simplify $\frac{2 {(0)}^{2} - 5 \cdot 0 - 12}{(0) - 4}$ .

Simplify the numerator.

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

$y = \frac{2 \cdot 0 - 5 \cdot 0 - 12}{0 - 4}$

Multiply $2$ by $0$ .

$y = \frac{0 - 5 \cdot 0 - 12}{0 - 4}$

Multiply $- 5$ by $0$ .

$y = \frac{0 + 0 - 12}{0 - 4}$

Add $0$ and $0$ .

$y = \frac{0 - 12}{0 - 4}$

Subtract $12$ from $0$ .

$y = \frac{- 12}{0 - 4}$

Simplify the expression.

Subtract $4$ from $0$ .

$y = \frac{- 12}{- 4}$

Divide $- 12$ by $- 4$ .

$y = 3$

y-intercept(s) in point form.

y-intercept(s): $(0, 3)$

Step 4

List the intersections.

x-intercept(s): $(- \frac{3}{2}, 0)$

y-intercept(s): $(0, 3)$

Step 5

Find the x and y Intercepts (2x^2-5x-12)/(x-4)

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