Question

Find the vertex of the function given below. Y=3x^2+6x+1

Answer 1

Answer:

B

Step-by-step explanation:

Given 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 + 6x + 1 ← is in standard form

with a = 3 and b = 6 , then

$x_{vertex}$ = - $\frac{6}{6}$ = - 1

Substitute x = - 1 into y for corresponding y- coordinate

y = 3(- 1)² + 6(- 1) + 1

  = 3(1) - 6 + 1

  = 3 - 5

  = - 2

vertex = (- 1, - 2 )