What are the coordinates of the vertex for f(x) = 3x2 + 6x + 3? (−1, 0) (3, 6) (0, −1) (6, 3)

for
ax^2+bx+c=0
x value of vertex is -b/2a
y value is x value subsituted

3x^2+6x+3
a=3
b=6
-6/(2*3)=-6/6=-1

x coordinate is -1
plug it in
f(-1)=3(-1)^2+6(-1)+3
f(-1)=3(1)-6+3
f(-1)=0

the vertex is at (-1,0)

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calculista calculista · Answer 2 · 2018-11-27T15:55:27+01:00

Answer:

The vertex is the point [tex](-1,0)[/tex]

Step-by-step explanation:

we have

[tex]f(x)=3x^{2} +6x+3[/tex]

we know ow that

The equation of a vertical parabola into vertex form is equal to

[tex]y=a(x-h)^{2}+k[/tex]

where

(h,k) is the vertex of the parabola

Convert the equation into vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

[tex]f(x)-3=3x^{2} +6x[/tex]

Factor the leading coefficient

[tex]f(x)-3=3(x^{2} +2x)[/tex]

Complete the square. Remember to balance the equation by adding the same constants to each side

[tex]f(x)-3+3=3(x^{2} +2x+1)[/tex]

[tex]f(x)=3(x^{2} +2x+1)[/tex]

Rewrite as perfect squares

[tex]f(x)=3(x+1)^{2}[/tex]

therefore

the vertex is the point [tex](-1,0)[/tex]