Graph x^2

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Algebra

Graph x^2

$x^{2}$

Find the properties of the given parabola.

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Rewrite the equation in vertex form.

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

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

$a = 1, b = 0, c = 0$

Consider the vertex form of a parabola.

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

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

$d = \frac{0}{2 (1)}$

Cancel the common factor of $0$ and $2$ .

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Factor $2$ out of $0$ .

$d = \frac{2 (0)}{2 (1)}$

Cancel the common factors.

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Cancel the common factor.

$d = \frac{2 \cdot 0}{2 \cdot 1}$

Rewrite the expression.

$d = \frac{0}{1}$

Divide $0$ by $1$ .

$d = 0$

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

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Simplify each term.

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

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

Multiply $4$ by $1$ .

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

Divide $0$ by $4$ .

$e = 0 - 0$

Multiply $- 1$ by $0$ .

$e = 0 + 0$

Add $0$ and $0$ .

$e = 0$

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

$x^{2}$

Set $y$ equal to the new right side.

$y = x^{2}$

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

$a = 1$

$h = 0$

$k = 0$

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

Opens Up

Find the vertex $(h, k)$ .

$(0, 0)$

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

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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 $1$ .

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Cancel the common factor.

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

Rewrite the expression.

$\frac{1}{4}$

Find the focus.

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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.

$(0, \frac{1}{4})$

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

$x = 0$

Find the directrix.

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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.

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

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

Direction: Opens Up

Vertex: $(0, 0)$

Focus: $(0, \frac{1}{4})$

Axis of Symmetry: $x = 0$

Directrix: $y = - \frac{1}{4}$

Direction: Opens Up

Vertex: $(0, 0)$

Focus: $(0, \frac{1}{4})$

Axis of Symmetry: $x = 0$

Directrix: $y = - \frac{1}{4}$

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.

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

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

Simplify the result.

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

$f (- 1) = 1$

The final answer is $1$ .

$1$

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

$y = 1$

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

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

Simplify the result.

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

$f (- 2) = 4$

The final answer is $4$ .

$4$

The $y$ value at $x = - 2$ is $4$ .

$y = 4$

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

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

Simplify the result.

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One to any power is one.

$f (1) = 1$

The final answer is $1$ .

$1$

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

$y = 1$

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

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

Simplify the result.

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

$f (2) = 4$

The final answer is $4$ .

$4$

The $y$ value at $x = 2$ is $4$ .

$y = 4$

Graph the parabola using its properties and the selected points.

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

Graph the parabola using its properties and the selected points.

Direction: Opens Up

Vertex: $(0, 0)$

Focus: $(0, \frac{1}{4})$

Axis of Symmetry: $x = 0$

Directrix: $y = - \frac{1}{4}$

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

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