What is the vertex of the following equation?

Question

y = 3x^2 – 12 x + 13

Answer 1

y = 3x²-12x+13 ---------------------> (1)

This eqation compared to y = ax²+bx+c

Than a = 3 , b = -12 and c=13

vertex line ordinate  x = (-b/2a)

Substitute the values a = 3 , b = -12 in point x.

x = -(-12/2(3))

x = -(-12/6)

Product of two same signs is positive

x = 12/6

x =2

There fore x = 2

Substitute x = 2 in y = 3x²-12x+13.

y = 3(2)²-12(2)+13

Evaluate the power 2² = 4

y = 3(4) -12(2) + 13

Multiply the each terms

y = 12-24+13

Simplify

y = 1

vertex coordinates (x, y) = (2, 1)

Answer 2

The equation is y = 3x^2 - 12x + 13.

y = 3(x^2 - 4x ) + 13

Add 4 inside the bracket in-order to complete the square. Hence subtract 12 outside the bracket.

y = 3(x^2 - 4x + 4) - 12 + 13

y = 3(x - 2)^2 - 12 + 13

y = 3(x - 2)^2 + 1

Compare the above equation with the parabolic equation y = a(x - h )^2 + k where vertex = (h, k ).

Therefore vertex (h,k ) = (2,1)