Find the x and y Intercepts f(x)=- square root of x+1-3

$f (x) = - \sqrt{x + 1} - 3$

Find the x-intercepts.

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

$0 = - \sqrt{x + 1} - 3$

Solve the equation.

Rewrite the equation as $- \sqrt{x + 1} - 3 = 0$ .

$- \sqrt{x + 1} - 3 = 0$

Add $3$ to both sides of the equation.

$- \sqrt{x + 1} = 3$

To remove the radical on the left side of the equation, square both sides of the equation.

${(- \sqrt{x + 1})}^{2} = 3^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x + 1}$ as ${(x + 1)}^{\frac{1}{2}}$ .

${(- {(x + 1)}^{\frac{1}{2}})}^{2} = 3^{2}$

Simplify the left side.

Simplify ${(- {(x + 1)}^{\frac{1}{2}})}^{2}$ .

Apply the product rule to $- {(x + 1)}^{\frac{1}{2}}$ .

${(- 1)}^{2} {({(x + 1)}^{\frac{1}{2}})}^{2} = 3^{2}$

Raise $- 1$ to the power of $2$ .

$1 {({(x + 1)}^{\frac{1}{2}})}^{2} = 3^{2}$

Multiply ${({(x + 1)}^{\frac{1}{2}})}^{2}$ by $1$ .

${({(x + 1)}^{\frac{1}{2}})}^{2} = 3^{2}$

Multiply the exponents in ${({(x + 1)}^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(x + 1)}^{\frac{1}{2} \cdot 2} = 3^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(x + 1)}^{\frac{1}{2} \cdot 2} = 3^{2}$

Rewrite the expression.

${(x + 1)}^{1} = 3^{2}$

Simplify.

$x + 1 = 3^{2}$

Simplify the right side.

Raise $3$ to the power of $2$ .

$x + 1 = 9$

Move all terms not containing $x$ to the right side of the equation.

Subtract $1$ from both sides of the equation.

$x = 9 - 1$

Subtract $1$ from $9$ .

$x = 8$

Exclude the solutions that do not make $0 = - \sqrt{x + 1} - 3$ true.

$No solution$

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

x-intercept(s): $None$

Find the y-intercepts.

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

$y = - \sqrt{(0) + 1} - 3$

Simplify $- \sqrt{(0) + 1} - 3$ .

Simplify each term.

Add $0$ and $1$ .

$y = - \sqrt{1} - 3$

Any root of $1$ is $1$ .

$y = - 1 \cdot 1 - 3$

Multiply $- 1$ by $1$ .

$y = - 1 - 3$

Subtract $3$ from $- 1$ .

$y = - 4$

y-intercept(s) in point form.

y-intercept(s): $(0, - 4)$

List the intersections.

x-intercept(s): $None$

y-intercept(s): $(0, - 4)$

Find the x and y Intercepts f(x)=- square root of x+1-3

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