Graph y=|x^2+2x-3|

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

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

For each $x$ value, there is one $y$ value. Select a few $x$ values from the domain. It would be more useful to select the values so that they are around the $x$ value of the absolute value vertex.

Substitute the $x$ value $- 2$ into $f (x) = | x^{2} + 2 x - 3 |$ . In this case, the point is $(- 2, 3)$ .

Replace the variable $x$ with $- 2$ in the expression.

$f (- 2) = | {(- 2)}^{2} + 2 (- 2) - 3 |$

Simplify the result.

Simplify each term.

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

$f (- 2) = | 4 + 2 (- 2) - 3 |$

Multiply $2$ by $- 2$ .

$f (- 2) = | 4 - 4 - 3 |$

Simplify by subtracting numbers.

Subtract $4$ from $4$ .

$f (- 2) = | 0 - 3 |$

Subtract $3$ from $0$ .

$f (- 2) = | - 3 |$

The absolute value is the distance between a number and zero. The distance between $- 3$ and $0$ is $3$ .

$f (- 2) = 3$

The final answer is $3$ .

$y = 3$

Substitute the $x$ value $- 1$ into $f (x) = | x^{2} + 2 x - 3 |$ . In this case, the point is $(- 1, 4)$ .

Replace the variable $x$ with $- 1$ in the expression.

$f (- 1) = | {(- 1)}^{2} + 2 (- 1) - 3 |$

Simplify the result.

Simplify each term.

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

$f (- 1) = | 1 + 2 (- 1) - 3 |$

Multiply $2$ by $- 1$ .

$f (- 1) = | 1 - 2 - 3 |$

Simplify by subtracting numbers.

Subtract $2$ from $1$ .

$f (- 1) = | - 1 - 3 |$

Subtract $3$ from $- 1$ .

$f (- 1) = | - 4 |$

The absolute value is the distance between a number and zero. The distance between $- 4$ and $0$ is $4$ .

$f (- 1) = 4$

The final answer is $4$ .

$y = 4$

Substitute the $x$ value $0$ into $f (x) = | x^{2} + 2 x - 3 |$ . In this case, the point is $(0, 3)$ .

Replace the variable $x$ with $0$ in the expression.

$f (0) = | {(0)}^{2} + 2 (0) - 3 |$

Simplify the result.

Simplify each term.

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

$f (0) = | 0 + 2 (0) - 3 |$

Multiply $2$ by $0$ .

$f (0) = | 0 + 0 - 3 |$

Simplify by adding and subtracting.

Add $0$ and $0$ .

$f (0) = | 0 - 3 |$

Subtract $3$ from $0$ .

$f (0) = | - 3 |$

The absolute value is the distance between a number and zero. The distance between $- 3$ and $0$ is $3$ .

$f (0) = 3$

The final answer is $3$ .

$y = 3$

Substitute the $x$ value $1$ into $f (x) = | x^{2} + 2 x - 3 |$ . In this case, the point is $(1, 0)$ .

Replace the variable $x$ with $1$ in the expression.

$f (1) = | {(1)}^{2} + 2 (1) - 3 |$

Simplify the result.

Simplify each term.

One to any power is one.

$f (1) = | 1 + 2 (1) - 3 |$

Multiply $2$ by $1$ .

$f (1) = | 1 + 2 - 3 |$

Simplify by adding and subtracting.

Add $1$ and $2$ .

$f (1) = | 3 - 3 |$

Subtract $3$ from $3$ .

$f (1) = | 0 |$

The absolute value is the distance between a number and zero. The distance between $0$ and $0$ is $0$ .

$f (1) = 0$

The final answer is $0$ .

$y = 0$

The absolute value can be graphed using the points around the vertex $(- 2, 3), (- 1, 4), (0, 3), (1, 0)$

$\begin{array}{c} x & y \\ - 2 & 3 \\ - 1 & 4 \\ 0 & 3 \\ 1 & 0 \end{array}$

Graph y=|x^2+2x-3|

Related Questions

Support LCMGCF.com by disabling your adblocker.