Precalculus Examples

h (x) = - x^{2} - 2 x + 3

$h (x) = - x^{2} - 2 x + 3$

Step 1

Find the properties of the given parabola.

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Step 1.1

Rewrite the equation in vertex form.

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Step 1.1.1

Complete the square for

- x^{2} - 2 x + 3

$- x^{2} - 2 x + 3$ .

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Step 1.1.1.1

Use the form

a x^{2} + b x + c

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

a

$a$ ,

b

$b$ , and

c

$c$ .

a = - 1

$a = - 1$

b = - 2

$b = - 2$

c = 3

$c = 3$

Step 1.1.1.2

Consider the vertex form of a parabola.

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

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

Step 1.1.1.3

Find the value of

d

$d$ using the formula

d = \frac{b}{2 a}

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

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Step 1.1.1.3.1

Substitute the values of

a

$a$ and

b

$b$ into the formula

d = \frac{b}{2 a}

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

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

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

Step 1.1.1.3.2

Simplify the right side.

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Step 1.1.1.3.2.1

Cancel the common factor of

- 2

$- 2$ and

2

$2$ .

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Step 1.1.1.3.2.1.1

Factor

2

$2$ out of

- 2

$- 2$ .

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

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

Step 1.1.1.3.2.1.2

Move the negative one from the denominator of

\frac{- 1}{- 1}

$\frac{- 1}{- 1}$ .

d = - 1 \cdot - 1

$d = - 1 \cdot - 1$

d = - 1 \cdot - 1

$d = - 1 \cdot - 1$

Step 1.1.1.3.2.2

Rewrite

- 1 \cdot - 1

$- 1 \cdot - 1$ as

- - 1

$- - 1$ .

d = - - 1

$d = - - 1$

Step 1.1.1.3.2.3

Multiply

- 1

$- 1$ by

- 1

$- 1$ .

d = 1

$d = 1$

d = 1

$d = 1$

d = 1

$d = 1$

Step 1.1.1.4

Find the value of

e

$e$ using the formula

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

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

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Step 1.1.1.4.1

Substitute the values of

c

$c$ ,

b

$b$ and

a

$a$ into the formula

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

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

e = 3 - \frac{{(- 2)}^{2}}{4 \cdot - 1}

$e = 3 - \frac{{(- 2)}^{2}}{4 \cdot - 1}$

Step 1.1.1.4.2

Simplify the right side.

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Step 1.1.1.4.2.1

Simplify each term.

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Step 1.1.1.4.2.1.1

Raise

- 2

$- 2$ to the power of

2

$2$ .

e = 3 - \frac{4}{4 \cdot - 1}

$e = 3 - \frac{4}{4 \cdot - 1}$

Step 1.1.1.4.2.1.2

Multiply

4

$4$ by

- 1

$- 1$ .

e = 3 - \frac{4}{- 4}

$e = 3 - \frac{4}{- 4}$

Step 1.1.1.4.2.1.3

Divide

4

$4$ by

- 4

$- 4$ .

e = 3 - - 1

$e = 3 - - 1$

Step 1.1.1.4.2.1.4

Multiply

- 1

$- 1$ by

- 1

$- 1$ .

e = 3 + 1

$e = 3 + 1$

e = 3 + 1

$e = 3 + 1$

Step 1.1.1.4.2.2

Add

3

$3$ and

1

$1$ .

e = 4

$e = 4$

e = 4

$e = 4$

e = 4

$e = 4$

Step 1.1.1.5

Substitute the values of

a

$a$ ,

d

$d$ , and

e

$e$ into the vertex form

- {(x + 1)}^{2} + 4

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

- {(x + 1)}^{2} + 4

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

- {(x + 1)}^{2} + 4

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

Step 1.1.2

Set

y

$y$ equal to the new right side.

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

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

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

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

Step 1.2

Use the vertex form,

y = a {(x - h)}^{2} + k

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

a

$a$ ,

h

$h$ , and

k

$k$ .

a = - 1

$a = - 1$

h = - 1

$h = - 1$

k = 4

$k = 4$

Step 1.3

Since the value of

a

$a$ is negative, the parabola opens down.

Opens Down

Step 1.4

Find the vertex

$(h, k)$ .

$(- 1, 4)$

Step 1.5

Find

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

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Step 1.5.1

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

$\frac{1}{4 a}$

Step 1.5.2

Substitute the value of

$a$ into the formula.

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

Step 1.5.3

Cancel the common factor of

$1$ and

$- 1$ .

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Step 1.5.3.1

Rewrite

$1$ as

$- 1 (- 1)$ .

$\frac{- 1 (- 1)}{4 \cdot - 1}$

Step 1.5.3.2

Move the negative in front of the fraction.

$- \frac{1}{4}$

Step 1.6

Find the focus.

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Step 1.6.1

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 1.6.2

Substitute the known values of

$h$ ,

$p$ , and

$k$ into the formula and simplify.

$(- 1, \frac{15}{4})$

Step 1.7

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

$x = - 1$

Step 1.8

Find the directrix.

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Step 1.8.1

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 1.8.2

Substitute the known values of

$p$ and

$k$ into the formula and simplify.

$y = \frac{17}{4}$

Step 1.9

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

Direction: Opens Down

Vertex:

$(- 1, 4)$

Focus:

$(- 1, \frac{15}{4})$

Axis of Symmetry:

$x = - 1$

Directrix:

$y = \frac{17}{4}$

Direction: Opens Down

Vertex:

$(- 1, 4)$

Focus:

$(- 1, \frac{15}{4})$

Axis of Symmetry:

$x = - 1$

Directrix:

$y = \frac{17}{4}$

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.

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Step 2.1

Replace the variable

$x$ with

$- 2$ in the expression.

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

Step 2.2

Simplify the result.

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Step 2.2.1

Simplify each term.

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Step 2.2.1.1

Raise

$- 2$ to the power of

$2$ .

$f (- 2) = - 1 \cdot 4 - 2 \cdot - 2 + 3$

Step 2.2.1.2

Multiply

$- 1$ by

$4$ .

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

Step 2.2.1.3

Multiply

$- 2$ by

$- 2$ .

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

Step 2.2.2

Simplify by adding numbers.

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Step 2.2.2.1

Add

$- 4$ and

$4$ .

$f (- 2) = 0 + 3$

Step 2.2.2.2

Add

$0$ and

$3$ .

$f (- 2) = 3$

Step 2.2.3

The final answer is

$3$ .

$3$

Step 2.3

The

$y$ value at

$x = - 2$ is

$3$ .

$y = 3$

Step 2.4

Replace the variable

$x$ with

$- 3$ in the expression.

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

Step 2.5

Simplify the result.

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Step 2.5.1

Simplify each term.

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Step 2.5.1.1

Raise

$- 3$ to the power of

$2$ .

$f (- 3) = - 1 \cdot 9 - 2 \cdot - 3 + 3$

Step 2.5.1.2

Multiply

$- 1$ by

$9$ .

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

Step 2.5.1.3

Multiply

$- 2$ by

$- 3$ .

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

Step 2.5.2

Simplify by adding numbers.

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Step 2.5.2.1

Add

$- 9$ and

$6$ .

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

Step 2.5.2.2

Add

$- 3$ and

$3$ .

$f (- 3) = 0$

Step 2.5.3

The final answer is

$0$ .

$0$

Step 2.6

The

$y$ value at

$x = - 3$ is

$0$ .

$y = 0$

Step 2.7

Replace the variable

$x$ with

$0$ in the expression.

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

Step 2.8

Simplify the result.

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Step 2.8.1

Simplify each term.

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Step 2.8.1.1

Raising

$0$ to any positive power yields

$0$ .

$f (0) = - 0 - 2 \cdot 0 + 3$

Step 2.8.1.2

Multiply

$- 1$ by

$0$ .

$f (0) = 0 - 2 \cdot 0 + 3$

Step 2.8.1.3

Multiply

$- 2$ by

$0$ .

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

Step 2.8.2

Simplify by adding numbers.

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Step 2.8.2.1

Add

$0$ and

$0$ .

$f (0) = 0 + 3$

Step 2.8.2.2

Add

$0$ and

$3$ .

$f (0) = 3$

Step 2.8.3

The final answer is

$3$ .

$3$

Step 2.9

The

$y$ value at

$x = 0$ is

$3$ .

$y = 3$

Step 2.10

Replace the variable

$x$ with

$1$ in the expression.

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

Step 2.11

Simplify the result.

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Step 2.11.1

Simplify each term.

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Step 2.11.1.1

One to any power is one.

$f (1) = - 1 \cdot 1 - 2 \cdot 1 + 3$

Step 2.11.1.2

Multiply

$- 1$ by

$1$ .

$f (1) = - 1 - 2 \cdot 1 + 3$

Step 2.11.1.3

Multiply

$- 2$ by

$1$ .

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

Step 2.11.2

Simplify by adding and subtracting.

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Step 2.11.2.1

Subtract

$2$ from

$- 1$ .

$f (1) = - 3 + 3$

Step 2.11.2.2

Add

$- 3$ and

$3$ .

$f (1) = 0$

Step 2.11.3

The final answer is

$0$ .

$0$

Step 2.12

The

$y$ value at

$x = 1$ is

$0$ .

$y = 0$

Step 2.13

Graph the parabola using its properties and the selected points.

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

Step 3

Graph the parabola using its properties and the selected points.

Direction: Opens Down

Vertex:

$(- 1, 4)$

Focus:

$(- 1, \frac{15}{4})$

Axis of Symmetry:

$x = - 1$

Directrix:

$y = \frac{17}{4}$

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

Step 4