akemezaug5

2023-03-06

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

Hunter Hendricks

Beginner2023-03-07Added 6 answers

$\text{given the parabola in}\phantom{\rule{1ex}{0ex}}{\text{standard form}}$

$\u2022{x}y=a{x}^{2}+bx+c{x};a\ne 0$

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

$\u2022{x}{x}_{{\text{vertex}}}=-\frac{b}{2a}$

$f\left(x\right)={x}^{2}+6x+5\phantom{\rule{1ex}{0ex}}\text{is in standard form}$

$\text{with}\phantom{\rule{1ex}{0ex}}a=1,b=6\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}c=5$

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

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

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

$\Rightarrow \textcolor[rgb]{}{\text{vertex}\phantom{\rule{1ex}{0ex}}}=(-3,-4)$

graph{x^2+6x+5 [-10, 10, -5, 5]}

$\u2022{x}y=a{x}^{2}+bx+c{x};a\ne 0$

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

$\u2022{x}{x}_{{\text{vertex}}}=-\frac{b}{2a}$

$f\left(x\right)={x}^{2}+6x+5\phantom{\rule{1ex}{0ex}}\text{is in standard form}$

$\text{with}\phantom{\rule{1ex}{0ex}}a=1,b=6\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}c=5$

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

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

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

$\Rightarrow \textcolor[rgb]{}{\text{vertex}\phantom{\rule{1ex}{0ex}}}=(-3,-4)$

graph{x^2+6x+5 [-10, 10, -5, 5]}

Kiara Rollins

Beginner2023-03-08Added 6 answers

The given function:

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

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

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

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

The following equation is a vertical parabola in its standard form:

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

Which has vertex at

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

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

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

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

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

The following equation is a vertical parabola in its standard form:

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

Which has vertex at

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