Question

Find the axis of symmetry and the vertex of the graph of y = 3x2 + 2x.

Answer 1

Step-by-step explanation:

Finding Vertex

Given

$y\:=\:3x^2\:+\:2x$

The vertex of an up-down facing parabola of the form

$y=ax^2+bx+c\:\mathrm{is}\:x_v=-\frac{b}{2a}$

The Parabola params are:

$a=3,\:b=2,\:c=0$

$x_v=-\frac{b}{2a}$

$x_v=-\frac{2}{2\cdot \:3}$

$x_v=-\frac{1}{3}$

$\mathrm{Plug\:in}\:\:x_v=-\frac{1}{3}\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}$

$y_v=3\left(-\frac{1}{3}\right)^2+2\left(-\frac{1}{3}\right)$

    $=3\left(-\frac{1}{3}\right)^2-2\cdot \frac{1}{3}$

    $=\frac{1}{3}-\frac{2}{3}$         ∵  $3\left(-\frac{1}{3}\right)^2=\frac{1}{3}$

    $=\frac{1-2}{3}$

    $=\frac{-1}{3}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{b}=-\frac{a}{b}$

   $=-\frac{1}{3}$

$y_v=-\frac{1}{3}$

Therefore the parabola vertex is:

$\left(-\frac{1}{3},\:-\frac{1}{3}\right)$

$\mathrm{If}\:a$

$\mathrm{If}\:a>0,\:\mathrm{then\:the\:vertex\:is\:a\:minimum\:value}$

$a=3$

$\mathrm{Minimum}\space\left(-\frac{1}{3},\:-\frac{1}{3}\right)$

Finding symmetry

For a parabola in standard form $y=ax^2+bx+c$

the axis of symmetry is the vertical line that goes through the vertex $x=\frac{-b}{2a}$

$\mathrm{Axis\:of\:Symmetry\:for}\:y=ax^2+bx+c\:\mathrm{is}\:x=\frac{-b}{2a}$

$a=3,\:b=2$

$x=\frac{-2}{2\cdot \:3}$

$x=-\frac{1}{3}$

Therefore,

$\mathrm{Axis\:of\:Symmetry\:for}\:y=3x^2+2x:\quad x=-\frac{1}{3}$