Respuesta :
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