Algebra Examples

$y = - 2$ ${(x - 1)}^{2} - 5$

Step 1

Graph $y = - 2$ .

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Step 1.1

Use the slope-intercept form to find the slope and y-intercept.

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Step 1.1.1

The slope-intercept form is $y = m x + b$ , where $m$ is the slope and $b$ is the y-intercept.

$y = m x + b$

Step 1.1.2

Find the values of $m$ and $b$ using the form $y = m x + b$ .

$m = 0$

$b = - 2$

Step 1.1.3

The slope of the line is the value of $m$ , and the y-intercept is the value of $b$ .

Slope: $0$

y-intercept: $(0, - 2)$

Slope: $0$

y-intercept: $(0, - 2)$

Step 1.2

Find two points on the line.

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

Step 1.3

Graph the line using the slope, y-intercept, and two points.

Slope: $0$

y-intercept: $(0, - 2)$

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

Slope: $0$

y-intercept: $(0, - 2)$

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

Step 2

Graph ${(x - 1)}^{2} - 5$ .

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Step 2.1

Find the properties of the given parabola.

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Step 2.1.1

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

$a = 1$

$h = 1$

$k = - 5$

Step 2.1.2

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

Opens Up

Step 2.1.3

Find the vertex $(h, k)$ .

$(1, - 5)$

Step 2.1.4

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

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Step 2.1.4.1

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

$\frac{1}{4 a}$

Step 2.1.4.2

Substitute the value of $a$ into the formula.

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

Step 2.1.4.3

Cancel the common factor of $1$ .

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Step 2.1.4.3.1

Cancel the common factor.

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

Step 2.1.4.3.2

Rewrite the expression.

$\frac{1}{4}$

Step 2.1.5

Find the focus.

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Step 2.1.5.1

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

$(h, k + p)$

Step 2.1.5.2

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

$(1, - \frac{19}{4})$

Step 2.1.6

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

$x = 1$

Step 2.1.7

Find the directrix.

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Step 2.1.7.1

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.

$y = k - p$

Step 2.1.7.2

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

$y = - \frac{21}{4}$

Step 2.1.8

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

Direction: Opens Up

Vertex: $(1, - 5)$

Focus: $(1, - \frac{19}{4})$

Axis of Symmetry: $x = 1$

Directrix: $y = - \frac{21}{4}$

Direction: Opens Up

Vertex: $(1, - 5)$

Focus: $(1, - \frac{19}{4})$

Axis of Symmetry: $x = 1$

Directrix: $y = - \frac{21}{4}$

Step 2.2

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.

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Step 2.2.1

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

$f (0) = {(0)}^{2} - 2 \cdot 0 - 4$

Step 2.2.2

Simplify the result.

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Step 2.2.2.1

Simplify each term.

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Step 2.2.2.1.1

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

$f (0) = 0 - 2 \cdot 0 - 4$

Step 2.2.2.1.2

Multiply $- 2$ by $0$ .

$f (0) = 0 + 0 - 4$

Step 2.2.2.2

Simplify by adding and subtracting.

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Step 2.2.2.2.1

Add $0$ and $0$ .

$f (0) = 0 - 4$

Step 2.2.2.2.2

Subtract $4$ from $0$ .

$f (0) = - 4$

Step 2.2.2.3

The final answer is $- 4$ .

$- 4$

Step 2.2.3

The $y$ value at $x = 0$ is $- 4$ .

$y = - 4$

Step 2.2.4

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

$f (- 1) = {(- 1)}^{2} - 2 \cdot - 1 - 4$

Step 2.2.5

Simplify the result.

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Step 2.2.5.1

Simplify each term.

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Step 2.2.5.1.1

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

$f (- 1) = 1 - 2 \cdot - 1 - 4$

Step 2.2.5.1.2

Multiply $- 2$ by $- 1$ .

$f (- 1) = 1 + 2 - 4$

Step 2.2.5.2

Simplify by adding and subtracting.

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Step 2.2.5.2.1

Add $1$ and $2$ .

$f (- 1) = 3 - 4$

Step 2.2.5.2.2

Subtract $4$ from $3$ .

$f (- 1) = - 1$

Step 2.2.5.3

The final answer is $- 1$ .

$- 1$

Step 2.2.6

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

$y = - 1$

Step 2.2.7

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

$f (2) = {(2)}^{2} - 2 \cdot 2 - 4$

Step 2.2.8

Simplify the result.

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Step 2.2.8.1

Simplify each term.

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Step 2.2.8.1.1

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

$f (2) = 4 - 2 \cdot 2 - 4$

Step 2.2.8.1.2

Multiply $- 2$ by $2$ .

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

Step 2.2.8.2

Simplify by subtracting numbers.

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Step 2.2.8.2.1

Subtract $4$ from $4$ .

$f (2) = 0 - 4$

Step 2.2.8.2.2

Subtract $4$ from $0$ .

$f (2) = - 4$

Step 2.2.8.3

The final answer is $- 4$ .

$- 4$

Step 2.2.9

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

$y = - 4$

Step 2.2.10

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

$f (3) = {(3)}^{2} - 2 \cdot 3 - 4$

Step 2.2.11

Simplify the result.

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Step 2.2.11.1

Simplify each term.

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Step 2.2.11.1.1

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

$f (3) = 9 - 2 \cdot 3 - 4$

Step 2.2.11.1.2

Multiply $- 2$ by $3$ .

$f (3) = 9 - 6 - 4$

Step 2.2.11.2

Simplify by subtracting numbers.

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Step 2.2.11.2.1

Subtract $6$ from $9$ .

$f (3) = 3 - 4$

Step 2.2.11.2.2

Subtract $4$ from $3$ .

$f (3) = - 1$

Step 2.2.11.3

The final answer is $- 1$ .

$- 1$

Step 2.2.12

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

$y = - 1$

Step 2.2.13

Graph the parabola using its properties and the selected points.

$\begin{array}{c} x & y \\ - 1 & - 1 \\ 0 & - 4 \\ 1 & - 5 \\ 2 & - 4 \\ 3 & - 1 \end{array}$

Step 2.3

Graph the parabola using its properties and the selected points.

Direction: Opens Up

Vertex: $(1, - 5)$

Focus: $(1, - \frac{19}{4})$

Axis of Symmetry: $x = 1$

Directrix: $y = - \frac{21}{4}$

$\begin{array}{c} x & y \\ - 1 & - 1 \\ 0 & - 4 \\ 1 & - 5 \\ 2 & - 4 \\ 3 & - 1 \end{array}$

Direction: Opens Up

Vertex: $(1, - 5)$

Focus: $(1, - \frac{19}{4})$

Axis of Symmetry: $x = 1$

Directrix: $y = - \frac{21}{4}$

$\begin{array}{c} x & y \\ - 1 & - 1 \\ 0 & - 4 \\ 1 & - 5 \\ 2 & - 4 \\ 3 & - 1 \end{array}$

Step 3

Plot each graph on the same coordinate system.

$y = - 2$

${(x - 1)}^{2} - 5$

Step 4