Algebra Examples

$y = x^{2} + 2 x + 1$

Step 1

Rewrite the equation in vertex form.

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Step 1.1

Complete the square for $x^{2} + 2 x + 1$ .

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Step 1.1.1

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

$a = 1$

$b = 2$

$c = 1$

Step 1.1.2

Consider the vertex form of a parabola.

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

Step 1.1.3

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

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Step 1.1.3.1

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

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

Step 1.1.3.2

Cancel the common factor of $2$ .

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Step 1.1.3.2.1

Cancel the common factor.

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

Step 1.1.3.2.2

Rewrite the expression.

$d = 1$

Step 1.1.4

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

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Step 1.1.4.1

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

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

Step 1.1.4.2

Simplify the right side.

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Step 1.1.4.2.1

Simplify each term.

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Step 1.1.4.2.1.1

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

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

Step 1.1.4.2.1.2

Multiply $4$ by $1$ .

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

Step 1.1.4.2.1.3

Cancel the common factor of $4$ .

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Step 1.1.4.2.1.3.1

Cancel the common factor.

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

Step 1.1.4.2.1.3.2

Rewrite the expression.

$e = 1 - 1 \cdot 1$

Step 1.1.4.2.1.4

Multiply $- 1$ by $1$ .

$e = 1 - 1$

Step 1.1.4.2.2

Subtract $1$ from $1$ .

$e = 0$

Step 1.1.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x + 1)}^{2} + 0$ .

${(x + 1)}^{2} + 0$

Step 1.2

Set $y$ equal to the new right side.

$y = {(x + 1)}^{2} + 0$

Step 2

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

$a = 1$

$h = - 1$

$k = 0$

Step 3

Find the vertex $(h, k)$ .

$(- 1, 0)$

Step 4