# How do I find the vertex of f(x)=x^2+6x+5?

Jul 22, 2018

$\text{vertex } = \left(- 3 , - 4\right)$

#### Explanation:

$\text{given the parabola in "color(blue)"standard form}$

•color(white)(x)y=ax^2+bx+c color(white)(x);a!=0

$\text{then the x-coordinate of the vertex is}$

•color(white)(x)x_(color(red)"vertex")=-b/(2a)

$f \left(x\right) = {x}^{2} + 6 x + 5 \text{ is in standard form}$

$\text{with "a=1,b=6" and } c = 5$

${x}_{\text{vertex}} = - \frac{6}{2} = - 3$

$\text{substitute this value into the equation for y}$

${y}_{\text{vertex}} = {\left(- 3\right)}^{2} + 6 \left(- 3\right) + 5 = - 4$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(- 3 , - 4\right)$
graph{x^2+6x+5 [-10, 10, -5, 5]}

$\left(- 3 , - 4\right)$

#### Explanation:

The given function:

$f \left(x\right) = {x}^{2} + 6 x + 5$

$y = {x}^{2} + 2 \left(3\right) x + {3}^{2} - {3}^{2} + 5$

$y = {\left(x + 3\right)}^{2} - 4$

${\left(x + 3\right)}^{2} = \left(y + 4\right)$

Above equation is in the standard form of vertical parabola:

${\left(x - {x}_{1}\right)}^{2} = 4 a \left(y - {y}_{1}\right)$

Which has vertex at

$\left({x}_{1} , {y}_{1}\right) \setminus \equiv \left(- 3 , - 4\right)$