Graph f(x)=x(x+1)

$f (x) = x (x + 1)$

Find the properties of the given parabola.

Rewrite the equation in vertex form.

Simplify $x (x + 1)$ .

Apply the distributive property.

$y = x \cdot x + x \cdot 1$

Simplify the expression.

Multiply $x$ by $x$ .

$y = x^{2} + x \cdot 1$

Multiply $x$ by $1$ .

$y = x^{2} + x$

Complete the square for $x^{2} + x$ .

Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = 1$

$b = 1$

$c = 0$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{1}{2 \cdot 1}$

Cancel the common factor of $1$ .

Cancel the common factor.

$d = \frac{1}{2 \cdot 1}$

Rewrite the expression.

$d = \frac{1}{2}$

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = 0 - \frac{1^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

One to any power is one.

$e = 0 - \frac{1}{4 \cdot 1}$

Multiply $4$ by $1$ .

$e = 0 - \frac{1}{4}$

Subtract $\frac{1}{4}$ from $0$ .

$e = - \frac{1}{4}$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x + \frac{1}{2})}^{2} - \frac{1}{4}$ .

${(x + \frac{1}{2})}^{2} - \frac{1}{4}$

Set $y$ equal to the new right side.

$y = {(x + \frac{1}{2})}^{2} - \frac{1}{4}$

Use the vertex form, $y = a {(x - h)}^{2} + k$ , to determine the values of $a$ , $h$ , and $k$ .

$a = 1$

$h = - \frac{1}{2}$

$k = - \frac{1}{4}$

Since the value of $a$ is positive, the parabola opens up.

Opens Up

Find the vertex $(h, k)$ .

$(- \frac{1}{2}, - \frac{1}{4})$

Find $p$ , the distance from the vertex to the focus.

Find the distance from the vertex to a focus of the parabola by using the following formula.

$\frac{1}{4 a}$

Substitute the value of $a$ into the formula.

$\frac{1}{4 \cdot 1}$

Cancel the common factor of $1$ .

Cancel the common factor.

$\frac{1}{4 \cdot 1}$

Rewrite the expression.

$\frac{1}{4}$

Find the focus.

The focus of a parabola can be found by adding $p$ to the y-coordinate $k$ if the parabola opens up or down.

$(h, k + p)$

Substitute the known values of $h$ , $p$ , and $k$ into the formula and simplify.

$(- \frac{1}{2}, 0)$

Find the axis of symmetry by finding the line that passes through the vertex and the focus.

$x = - \frac{1}{2}$

Find the directrix.

The directrix of a parabola is the horizontal line found by subtracting $p$ from the y-coordinate $k$ of the vertex if the parabola opens up or down.

$y = k - p$

Substitute the known values of $p$ and $k$ into the formula and simplify.

$y = - \frac{1}{2}$

Use the properties of the parabola to analyze and graph the parabola.

Direction: Opens Up

Vertex: $(- \frac{1}{2}, - \frac{1}{4})$

Focus: $(- \frac{1}{2}, 0)$

Axis of Symmetry: $x = - \frac{1}{2}$

Directrix: $y = - \frac{1}{2}$

Direction: Opens Up

Vertex: $(- \frac{1}{2}, - \frac{1}{4})$

Focus: $(- \frac{1}{2}, 0)$

Axis of Symmetry: $x = - \frac{1}{2}$

Directrix: $y = - \frac{1}{2}$

Select a few $x$ values, and plug them into the equation to find the corresponding $y$ values. The $x$ values should be selected around the vertex.

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

$f (- 1) = {(- 1)}^{2} - 1$

Simplify the result.

Remove parentheses.

$f (- 1) = {(- 1)}^{2} - 1$

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

$f (- 1) = 1 - 1$

Subtract $1$ from $1$ .

$f (- 1) = 0$

The final answer is $0$ .

$0$

The $y$ value at $x = - 1$ is $0$ .

$y = 0$

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

$f (- 2) = {(- 2)}^{2} - 2$

Simplify the result.

Remove parentheses.

$f (- 2) = {(- 2)}^{2} - 2$

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

$f (- 2) = 4 - 2$

Subtract $2$ from $4$ .

$f (- 2) = 2$

The final answer is $2$ .

$2$

The $y$ value at $x = - 2$ is $2$ .

$y = 2$

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

$f (1) = {(1)}^{2} + 1$

Simplify the result.

Remove parentheses.

$f (1) = {(1)}^{2} + 1$

One to any power is one.

$f (1) = 1 + 1$

Add $1$ and $1$ .

$f (1) = 2$

The final answer is $2$ .

$2$

The $y$ value at $x = 1$ is $2$ .

$y = 2$

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

$f (2) = {(2)}^{2} + 2$

Simplify the result.

Remove parentheses.

$f (2) = {(2)}^{2} + 2$

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

$f (2) = 4 + 2$

Add $4$ and $2$ .

$f (2) = 6$

The final answer is $6$ .

$6$

The $y$ value at $x = 2$ is $6$ .

$y = 6$

Graph the parabola using its properties and the selected points.

$\begin{array}{c} x & y \\ - 2 & 2 \\ - 1 & 0 \\ - \frac{1}{2} & - \frac{1}{4} \\ 1 & 2 \\ 2 & 6 \end{array}$

Graph the parabola using its properties and the selected points.

Direction: Opens Up

Vertex: $(- \frac{1}{2}, - \frac{1}{4})$

Focus: $(- \frac{1}{2}, 0)$

Axis of Symmetry: $x = - \frac{1}{2}$

Directrix: $y = - \frac{1}{2}$

$\begin{array}{c} x & y \\ - 2 & 2 \\ - 1 & 0 \\ - \frac{1}{2} & - \frac{1}{4} \\ 1 & 2 \\ 2 & 6 \end{array}$

Graph f(x)=x(x+1)

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