Find the Domain and Range y^2=x^2+2x-3

$y^{2} = x^{2} + 2 x - 3$

Step 1

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

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

Step 2

Factor $x^{2} + 2 x - 3$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 3$ and whose sum is $2$ .

$- 1, 3$

Write the factored form using these integers.

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

Step 3

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.

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

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

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

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

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

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

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

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

Step 4

Set the radicand in $\sqrt{(x - 1) (x + 3)}$ greater than or equal to $0$ to find where the expression is defined.

$(x - 1) (x + 3) \geq 0$

Step 5

Solve for $x$ .

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

$x - 1 = 0$

$x + 3 = 0$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The final solution is all the values that make $(x - 1) (x + 3) \geq 0$ true.

$x = 1, - 3$

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 1$

$x > 1$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < - 3$ to see if it makes the inequality true.

Choose a value on the interval $x < - 3$ and see if this value makes the original inequality true.

$x = - 6$

Replace $x$ with $- 6$ in the original inequality.

$((- 6) - 1) ((- 6) + 3) \geq 0$

The left side $21$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 3 < x < 1$ to see if it makes the inequality true.

Choose a value on the interval $- 3 < x < 1$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

$((0) - 1) ((0) + 3) \geq 0$

The left side $- 3$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 1$ to see if it makes the inequality true.

Choose a value on the interval $x > 1$ and see if this value makes the original inequality true.

$x = 4$

Replace $x$ with $4$ in the original inequality.

$((4) - 1) ((4) + 3) \geq 0$

The left side $21$ is greater than the right side $0$ , which means that the given statement is always true.

True

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 3$ True

$- 3 < x < 1$ False

$x > 1$ True

$x < - 3$ True

$- 3 < x < 1$ False

$x > 1$ True

The solution consists of all of the true intervals.

$x \leq - 3$ or $x \geq 1$

Step 6

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(- \infty, - 3] \cup [1, \infty)$

Set-Builder Notation:

${x | x \leq - 3, x \geq 1}$

Step 7

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${y | y \in ℝ}$

Step 8

Determine the domain and range.

Domain: $(- \infty, - 3] \cup [1, \infty), {x | x \leq - 3, x \geq 1}$

Range: $(- \infty, \infty), {y | y \in ℝ}$

Step 9

Find the Domain and Range y^2=x^2+2x-3

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