Graph f(x)=2x+x^2

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

Step 1

Find the properties of the given parabola.

Rewrite the equation in vertex form.

Reorder $2 x$ and $x^{2}$ .

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

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

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

$a = 1$

$b = 2$

$c = 0$

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 $d = \frac{b}{2 a}$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

$d = 1$

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 $e = c - \frac{b^{2}}{4 a}$ .

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

Simplify the right side.

Simplify each term.

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

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

Multiply $4$ by $1$ .

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

Cancel the common factor of $4$ .

Cancel the common factor.

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

Rewrite the expression.

$e = 0 - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$e = 0 - 1$

Subtract $1$ from $0$ .

$e = - 1$

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

${(x + 1)}^{2} - 1$

Set $y$ equal to the new right side.

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

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

$a = 1$

$h = - 1$

$k = - 1$

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

Opens Up

Find the vertex $(h, k)$ .

$(- 1, - 1)$

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.

$(- 1, - \frac{3}{4})$

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

$x = - 1$

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.

$y = - \frac{5}{4}$

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

Direction: Opens Up

Vertex: $(- 1, - 1)$

Focus: $(- 1, - \frac{3}{4})$

Axis of Symmetry: $x = - 1$

Directrix: $y = - \frac{5}{4}$

Direction: Opens Up

Vertex: $(- 1, - 1)$

Focus: $(- 1, - \frac{3}{4})$

Axis of Symmetry: $x = - 1$

Directrix: $y = - \frac{5}{4}$

Step 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 $- 2$ in the expression.

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

Simplify the result.

Simplify each term.

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

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

Multiply $2$ by $- 2$ .

$f (- 2) = 4 - 4$

Subtract $4$ from $4$ .

$f (- 2) = 0$

The final answer is $0$ .

$0$

The $y$ value at $x = - 2$ is $0$ .

$y = 0$

Replace the variable $x$ with $- 3$ in the expression.

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

Simplify the result.

Simplify each term.

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

$f (- 3) = 9 + 2 (- 3)$

Multiply $2$ by $- 3$ .

$f (- 3) = 9 - 6$

Subtract $6$ from $9$ .

$f (- 3) = 3$

The final answer is $3$ .

$3$

The $y$ value at $x = - 3$ is $3$ .

$y = 3$

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

$f (0) = {(0)}^{2} + 2 (0)$

Simplify the result.

Simplify each term.

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

$f (0) = 0 + 2 (0)$

Multiply $2$ by $0$ .

$f (0) = 0 + 0$

Add $0$ and $0$ .

$f (0) = 0$

The final answer is $0$ .

$0$

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

$y = 0$

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

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

Simplify the result.

Simplify each term.

One to any power is one.

$f (1) = 1 + 2 (1)$

Multiply $2$ by $1$ .

$f (1) = 1 + 2$

Add $1$ and $2$ .

$f (1) = 3$

The final answer is $3$ .

$3$

The $y$ value at $x = 1$ is $3$ .

$y = 3$

Graph the parabola using its properties and the selected points.

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

Step 3

Graph the parabola using its properties and the selected points.

Direction: Opens Up

Vertex: $(- 1, - 1)$

Focus: $(- 1, - \frac{3}{4})$

Axis of Symmetry: $x = - 1$

Directrix: $y = - \frac{5}{4}$

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

Step 4

Graph f(x)=2x+x^2

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