akemezaug5

2023-03-06

How do I find the vertex of $f\left(x\right)={x}^{2}+6x+5$?

Hunter Hendricks

$\text{given the parabola in}\phantom{\rule{1ex}{0ex}}\text{standard form}$
$•xy=a{x}^{2}+bx+cx;a\ne 0$
$\text{then the x-coordinate of the vertex is}$
$•x{x}_{\text{vertex}}=-\frac{b}{2a}$
$f\left(x\right)={x}^{2}+6x+5\phantom{\rule{1ex}{0ex}}\text{is in standard form}$
$\text{with}\phantom{\rule{1ex}{0ex}}a=1,b=6\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}c=5$
${x}_{\text{vertex}}=-\frac{6}{2}=-3$
$\text{substitute this value into the equation for y}$
${y}_{\text{vertex}}={\left(-3\right)}^{2}+6\left(-3\right)+5=-4$
$⇒\text{vertex}\phantom{\rule{1ex}{0ex}}=\left(-3,-4\right)$
graph{x^2+6x+5 [-10, 10, -5, 5]}

Kiara Rollins

The given function:
$f\left(x\right)={x}^{2}+6x+5$
$y={x}^{2}+2\left(3\right)x+{3}^{2}-{3}^{2}+5$
$y={\left(x+3\right)}^{2}-4$
${\left(x+3\right)}^{2}=\left(y+4\right)$
The following equation is a vertical parabola in its standard form:
${\left(x-{x}_{1}\right)}^{2}=4a\left(y-{y}_{1}\right)$
Which has vertex at
$\left({x}_{1},{y}_{1}\right)\equiv \left(-3,-4\right)$

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