Find the x and y Intercepts (2x^2-98)/(x+4)

$\frac{2 x^{2} - 98}{x + 4}$

Write $\frac{2 x^{2} - 98}{x + 4}$ as an equation.

$y = \frac{2 x^{2} - 98}{x + 4}$

Find the x-intercepts.

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

$0 = \frac{2 x^{2} - 98}{x + 4}$

Solve the equation.

Set the numerator equal to zero.

$2 x^{2} - 98 = 0$

Solve the equation for $x$ .

Add $98$ to both sides of the equation.

$2 x^{2} = 98$

Divide each term in $2 x^{2} = 98$ by $2$ and simplify.

Divide each term in $2 x^{2} = 98$ by $2$ .

$\frac{2 x^{2}}{2} = \frac{98}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x^{2}}{2} = \frac{98}{2}$

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{98}{2}$

Simplify the right side.

Divide $98$ by $2$ .

$x^{2} = 49$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{49}$

Simplify $\pm \sqrt{49}$ .

Rewrite $49$ as $7^{2}$ .

$x = \pm \sqrt{7^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 7$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 7$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 7$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 7, - 7$

x-intercept(s) in point form.

x-intercept(s): $(7, 0), (- 7, 0)$

Find the y-intercepts.

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

$y = \frac{2 {(0)}^{2} - 98}{(0) + 4}$

Solve the equation.

Remove parentheses.

$y = \frac{2 \cdot 0^{2} - 98}{(0) + 4}$

Remove parentheses.

$y = \frac{2 \cdot 0^{2} - 98}{0 + 4}$

Remove parentheses.

$y = \frac{2 {(0)}^{2} - 98}{(0) + 4}$

Simplify $\frac{2 {(0)}^{2} - 98}{(0) + 4}$ .

Cancel the common factor of $2 {(0)}^{2} - 98$ and $(0) + 4$ .

Factor $2$ out of $2 {(0)}^{2}$ .

$y = \frac{2 ({(0)}^{2}) - 98}{(0) + 4}$

Factor $2$ out of $- 98$ .

$y = \frac{2 {(0)}^{2} + 2 \cdot - 49}{(0) + 4}$

Factor $2$ out of $2 {(0)}^{2} + 2 \cdot - 49$ .

$y = \frac{2 ({(0)}^{2} - 49)}{(0) + 4}$

Cancel the common factors.

Factor $2$ out of $0$ .

$y = \frac{2 ({(0)}^{2} - 49)}{2 (0) + 4}$

Factor $2$ out of $4$ .

$y = \frac{2 ({(0)}^{2} - 49)}{2 \cdot 0 + 2 \cdot 2}$

Factor $2$ out of $2 \cdot 0 + 2 \cdot 2$ .

$y = \frac{2 ({(0)}^{2} - 49)}{2 \cdot (0 + 2)}$

Cancel the common factor.

$y = \frac{2 ({(0)}^{2} - 49)}{2 \cdot (0 + 2)}$

Rewrite the expression.

$y = \frac{{(0)}^{2} - 49}{0 + 2}$

Simplify the numerator.

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

$y = \frac{0 - 49}{0 + 2}$

Subtract $49$ from $0$ .

$y = \frac{- 49}{0 + 2}$

Simplify the expression.

Add $0$ and $2$ .

$y = \frac{- 49}{2}$

Move the negative in front of the fraction.

$y = - \frac{49}{2}$

y-intercept(s) in point form.

y-intercept(s): $(0, - \frac{49}{2})$

List the intersections.

x-intercept(s): $(7, 0), (- 7, 0)$

y-intercept(s): $(0, - \frac{49}{2})$

Find the x and y Intercepts (2x^2-98)/(x+4)

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