Examples

f (x) = x^{2} - 10 x + 25

$f (x) = x^{2} - 10 x + 25$

Step 1

Write

f (x) = x^{2} - 10 x + 25

$f (x) = x^{2} - 10 x + 25$ as an equation.

y = x^{2} - 10 x + 25

$y = x^{2} - 10 x + 25$

Step 2

Rewrite the equation in vertex form.

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Step 2.1

Complete the square for

x^{2} - 10 x + 25

$x^{2} - 10 x + 25$ .

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Step 2.1.1

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 = - 10

$b = - 10$

c = 25

$c = 25$

Step 2.1.2

Consider the vertex form of a parabola.

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

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

Step 2.1.3

Find the value of

d

$d$ using the formula

d = \frac{b}{2 a}

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

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Step 2.1.3.1

Substitute the values of

a

$a$ and

b

$b$ into the formula

d = \frac{b}{2 a}

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

d = \frac{- 10}{2 \cdot 1}

$d = \frac{- 10}{2 \cdot 1}$

Step 2.1.3.2

Cancel the common factor of

- 10

$- 10$ and

2

$2$ .

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Step 2.1.3.2.1

Factor

2

$2$ out of

- 10

$- 10$ .

d = \frac{2 \cdot - 5}{2 \cdot 1}

$d = \frac{2 \cdot - 5}{2 \cdot 1}$

Step 2.1.3.2.2

Cancel the common factors.

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Step 2.1.3.2.2.1

Factor

2

$2$ out of

2 \cdot 1

$2 \cdot 1$ .

d = \frac{2 \cdot - 5}{2 (1)}

$d = \frac{2 \cdot - 5}{2 (1)}$

Step 2.1.3.2.2.2

Cancel the common factor.

$d = \frac{2 \cdot - 5}{2 \cdot 1}$

Step 2.1.3.2.2.3

Rewrite the expression.

$d = \frac{- 5}{1}$

Step 2.1.3.2.2.4

Divide

$- 5$ by

$1$ .

$d = - 5$

Step 2.1.4

Find the value of

$e$ using the formula

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

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Step 2.1.4.1

Substitute the values of

$c$ ,

$b$ and

$a$ into the formula

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

$e = 25 - \frac{{(- 10)}^{2}}{4 \cdot 1}$

Step 2.1.4.2

Simplify the right side.

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Step 2.1.4.2.1

Simplify each term.

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Step 2.1.4.2.1.1

Raise

$- 10$ to the power of

$2$ .

$e = 25 - \frac{100}{4 \cdot 1}$

Step 2.1.4.2.1.2

Multiply

$4$ by

$1$ .

$e = 25 - \frac{100}{4}$

Step 2.1.4.2.1.3

Divide

$100$ by

$4$ .

$e = 25 - 1 \cdot 25$

Step 2.1.4.2.1.4

Multiply

$- 1$ by

$25$ .

$e = 25 - 25$

Step 2.1.4.2.2

Subtract

$25$ from

$25$ .

$e = 0$

Step 2.1.5

Substitute the values of

$a$ ,

$d$ , and

$e$ into the vertex form

${(x - 5)}^{2} + 0$ .

${(x - 5)}^{2} + 0$

Step 2.2

Set

$y$ equal to the new right side.

$y = {(x - 5)}^{2} + 0$

Step 3

Use the vertex form,

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

$a$ ,

$h$ , and

$k$ .

$a = 1$

$h = 5$

$k = 0$

Step 4

Since the value of

$a$ is positive, the parabola opens up.

Opens Up

Step 5

Find the vertex

$(h, k)$ .

$(5, 0)$

Step 6

Find

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

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Step 6.1

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

$\frac{1}{4 a}$

Step 6.2

Substitute the value of

$a$ into the formula.

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

Step 6.3

Cancel the common factor of

$1$ .

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Step 6.3.1

Cancel the common factor.

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

Step 6.3.2

Rewrite the expression.

$\frac{1}{4}$

Step 7

Find the focus.

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Step 7.1

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 7.2

Substitute the known values of

$h$ ,

$p$ , and

$k$ into the formula and simplify.

$(5, \frac{1}{4})$

Step 8

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

$x = 5$

Step 9

Find the directrix.

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Step 9.1

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 9.2

Substitute the known values of

$p$ and

$k$ into the formula and simplify.

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

Step 10

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

Direction: Opens Up

Vertex:

$(5, 0)$

Focus:

$(5, \frac{1}{4})$

Axis of Symmetry:

$x = 5$

Directrix:

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

Step 11

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