Find the x and y Intercepts y=(x-3)^2-4

$y = {(x - 3)}^{2} - 4$

Find the x-intercepts.

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

$0 = {(x - 3)}^{2} - 4$

Solve the equation.

Rewrite the equation as ${(x - 3)}^{2} - 4 = 0$ .

${(x - 3)}^{2} - 4 = 0$

Add $4$ to both sides of the equation.

${(x - 3)}^{2} = 4$

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

$x - 3 = \pm \sqrt{4}$

Simplify $\pm \sqrt{4}$ .

Rewrite $4$ as $2^{2}$ .

$x - 3 = \pm \sqrt{2^{2}}$

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

$x - 3 = \pm 2$

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.

$x - 3 = 2$

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

Add $3$ to both sides of the equation.

$x = 2 + 3$

Add $2$ and $3$ .

$x = 5$

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

$x - 3 = - 2$

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

Add $3$ to both sides of the equation.

$x = - 2 + 3$

Add $- 2$ and $3$ .

$x = 1$

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

$x = 5, 1$

x-intercept(s) in point form.

x-intercept(s): $(5, 0), (1, 0)$

Find the y-intercepts.

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

$y = {((0) - 3)}^{2} - 4$

Solve the equation.

Remove parentheses.

$y = {(0 - 3)}^{2} - 4$

Remove parentheses.

$y = {((0) - 3)}^{2} - 4$

Simplify ${((0) - 3)}^{2} - 4$ .

Simplify each term.

Subtract $3$ from $0$ .

$y = {(- 3)}^{2} - 4$

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

$y = 9 - 4$

Subtract $4$ from $9$ .

$y = 5$

y-intercept(s) in point form.

y-intercept(s): $(0, 5)$

List the intersections.

x-intercept(s): $(5, 0), (1, 0)$

y-intercept(s): $(0, 5)$

Find the x and y Intercepts y=(x-3)^2-4

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