Write Y=3x^2+12x+11 in form a(x-h)^2+k?

Question

Please explain

Answer 1

The equation y = 3x ² + 12x + 11

Vertex form of a parabola is y  = a  (x - h ) ² + k.

Here x ² coefficient is 3, for perfect square make x ² coefficient 1 by dividing each side by 3.

y  = 3x ² + 12x + 11

y /3 = x ² + 4x + 11/3

To change the expression into a perfect square trinomial add (half the x coefficient)² to each side of the expression.

x  coefficient = 4 then (half the x coefficient)² is (2) ² = 4.

So, add 4 to each side.

y /3 + 4 = x ²+ 4x + 11/3 + 4

y /3 + 4 = x ² + 4x + 4 + 11/3

y /3 + 4 = (x + 2) ² + 11/3

y /3 = (x + 2) ² +11/3 - 4

y /3 = (x + 2) ² - 1/3

y  = 3(x + 2)² - 1.

Therefore, vertex form of the parabola is  y  = a  (x - h ) ² + k.

y = 3[x - (-2)] ² - 1

a  = 3 , (h ,k ) = (-2 , -1).