# What is the vertex of y= x^2+4x + 1 ?

Nov 26, 2015

${P}_{\text{vertex}} = \left(- 2 , - 3\right)$

#### Explanation:

Given: $\textcolor{b r o w n}{y = {x}^{2} + 4 x + 1} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(1\right)$

Let point of vertex be ${P}_{\text{vertex}}$

Extract the 4 from $4 x$

Do the following to it:

$- \frac{1}{2} \times 4 = - 2$

${x}_{\text{vertex}} = \textcolor{b l u e}{- 2} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(2\right)$
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Substitute (2) into equation (1) to find ${y}_{\text{vertex}}$

$\textcolor{b r o w n}{{y}_{\text{vertex}} = {\textcolor{b l u e}{\left(- 2\right)}}^{2} + 4 \textcolor{b l u e}{\left(- 2\right)} + 1}$

${y}_{\text{vertex}} = 4 - 8 + 1 = - 3$
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