Question

Pls help quickly!!!

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

Answer 1

Answer:

vertex = ( $\frac{2}{3}$, $\frac{8}{3}$ )

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

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

with a = 3 , b = - 4 , then

x = - $\frac{-4}{6}$ = $\frac{2}{3}$

substitute x = $\frac{2}{3}$ into the equation for corresponding y- coordinate of vertex

y = 3($\frac{2}{3}$ )² - 4($\frac{2}{3}$ ) + 3

  = 3($\frac{4}{9}$ ) - $\frac{8}{3}$ + 3

  = $\frac{4}{3}$ - $\frac{8}{3}$ + $\frac{12}{3}$

  = $\frac{8}{3}$

vertex = ( $\frac{2}{3}$, $\frac{8}{3}$ )

Answer 2

Vertex ( 2/3 , 8/3 )