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Step-by-step explanation:

Finding Vertex

Given

[tex]y\:=\:3x^2\:+\:2x[/tex]

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

[tex]y=ax^2+bx+c\:\mathrm{is}\:x_v=-\frac{b}{2a}[/tex]

The Parabola params are:

[tex]a=3,\:b=2,\:c=0[/tex]

[tex]x_v=-\frac{b}{2a}[/tex]

[tex]x_v=-\frac{2}{2\cdot \:3}[/tex]

[tex]x_v=-\frac{1}{3}[/tex]

[tex]\mathrm{Plug\:in}\:\:x_v=-\frac{1}{3}\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}[/tex]

[tex]y_v=3\left(-\frac{1}{3}\right)^2+2\left(-\frac{1}{3}\right)[/tex]

    [tex]=3\left(-\frac{1}{3}\right)^2-2\cdot \frac{1}{3}[/tex]

    [tex]=\frac{1}{3}-\frac{2}{3}[/tex]         ∵  [tex]3\left(-\frac{1}{3}\right)^2=\frac{1}{3}[/tex]

    [tex]=\frac{1-2}{3}[/tex]

    [tex]=\frac{-1}{3}[/tex]

[tex]\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{b}=-\frac{a}{b}[/tex]

   [tex]=-\frac{1}{3}[/tex]

[tex]y_v=-\frac{1}{3}[/tex]

Therefore the parabola vertex is:

[tex]\left(-\frac{1}{3},\:-\frac{1}{3}\right)[/tex]

[tex]\mathrm{If}\:a<0,\:\mathrm{then\:the\:vertex\:is\:a\:maximum\:value}[/tex]

[tex]\mathrm{If}\:a>0,\:\mathrm{then\:the\:vertex\:is\:a\:minimum\:value}[/tex]

[tex]a=3[/tex]

[tex]\mathrm{Minimum}\space\left(-\frac{1}{3},\:-\frac{1}{3}\right)[/tex]

Finding symmetry

For a parabola in standard form [tex]y=ax^2+bx+c[/tex]

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

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

[tex]a=3,\:b=2[/tex]

[tex]x=\frac{-2}{2\cdot \:3}[/tex]

[tex]x=-\frac{1}{3}[/tex]

Therefore,

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

The equation [tex]y = 3x^2 + 2x[/tex] is an illustration of a quadratic function

The vertex (x,y) is [tex](-\frac 13,-\frac 13)[/tex], and the axis of symmetry is [tex]x = -\frac 13[/tex]

How to determine the vertex?

The equation of the function is given as:

[tex]y = 3x^2 + 2x[/tex]

Differentiate

[tex]y' = 6x + 2[/tex]

Set to 0

[tex]6x + 2 = 0[/tex]

Subtract 2 from both sides

[tex]6x = -2[/tex]

Divide by 6

[tex]x = -\frac 13[/tex]

Substitute -1/3 for x in [tex]y = 3x^2 + 2x[/tex]

[tex]y = 3(-\frac 13)^2 + 2(-\frac 13)[/tex]

Evaluate

[tex]y = \frac 13 -\frac 23[/tex]

This gives

[tex]y = -\frac 13[/tex]

Hence, the vertex (x,y) is [tex](-\frac 13,-\frac 13)[/tex]

The axis of symmetry

In (a), we have:

[tex]x = -\frac 13[/tex]

Hence, the axis of symmetry is [tex]x = -\frac 13[/tex]

Read more about quadratic functions at:

https://brainly.com/question/1349456

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