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Pls help quickly!!!

Find the coordinates of the vertex of the graph of the function.
y=3x^2-4x+3


Sagot :

Answer:

vertex = ( [tex]\frac{2}{3}[/tex], [tex]\frac{8}{3}[/tex] )

Step-by-step explanation:

given the equation of a parabola in standard form

y = ax² + bx + c ( a ≠ 0 )

then the x- coordinate of the vertex is

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

y = 3x² - 4x + 3 ← is in standard form

with a = 3 , b = - 4 , then

x = - [tex]\frac{-4}{6}[/tex] = [tex]\frac{2}{3}[/tex]

substitute x = [tex]\frac{2}{3}[/tex] into the equation for corresponding y- coordinate of vertex

y = 3([tex]\frac{2}{3}[/tex] )² - 4([tex]\frac{2}{3}[/tex] ) + 3

  = 3([tex]\frac{4}{9}[/tex] ) - [tex]\frac{8}{3}[/tex] + 3

  = [tex]\frac{4}{3}[/tex] - [tex]\frac{8}{3}[/tex] + [tex]\frac{12}{3}[/tex]

  = [tex]\frac{8}{3}[/tex]

vertex = ( [tex]\frac{2}{3}[/tex], [tex]\frac{8}{3}[/tex] )

Vertex ( 2/3 , 8/3 )