Find the Range y=x^2+4x+6

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Trigonometry

$y = x^{2} + 4 x + 6$

Locate the vertex of the parabola.

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Rewrite the equation in vertex form.

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Complete the square for $x^{2} + 4 x + 6$ .

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

$a = 1, b = 4, c = 6$

Consider the vertex form of a parabola.

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

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

$d = \frac{4}{2 (1)}$

Cancel the common factor of $4$ and $2$ .

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Factor $2$ out of $4$ .

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

Cancel the common factors.

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Factor $2$ out of $2 \cdot 1$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $2$ by $1$ .

$d = 2$

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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Cancel the common factor of ${(4)}^{2}$ and $4$ .

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Factor $4$ out of ${(4)}^{2}$ .

$e = 6 - \frac{4 \cdot 4}{4 (1)}$

Cancel the common factors.

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Cancel the common factor.

$e = 6 - \frac{4 \cdot 4}{4 \cdot 1}$

Rewrite the expression.

$e = 6 - \frac{4}{1}$

Divide $4$ by $1$ .

$e = 6 - 1 \cdot 4$

Multiply $- 1$ by $4$ .

$e = 6 - 4$

Subtract $4$ from $6$ .

$e = 2$

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

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

Set $y$ equal to the new right side.

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

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

$a = 1$

$h = - 2$

$k = 2$

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

Opens Up

Find the vertex $(h, k)$ .

$(- 2, 2)$

The range of a parabola that opens up starts at its vertex $(- 2, 2)$ and extends to infinity.

Interval Notation:

$[2, \infty)$

Set-Builder Notation:

${y | y \geq 2}$

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