What is the minimum value of $y$ if $y=3x^2+6x+9?$

Answer 1

The standard dorm of a auadratic equation:

y = a x² + b x + c

The vertex of a quadratic equation is either a maximum or a minimum of the function.

If leading coefficient a > 0, then the parabola opens up and vertex ts minimum.

If leading coefficient a < 0, then the parabola opens down and vertex ts maximum.

The value x = − b / 2a tells you the x value of the vertex.

In this case y = 3 x² + 6 x + 9

a = 3 , b = 6 , c = 9

a > 0, vertex is minimum

x = − b / 2a = - 6 / 2 ∙ 3 = - 6 / 6 = - 1

ymin = y(-1) = 3 ∙ ( - 1 )² + 6 ∙ ( - 1 ) + 9 = 3 ∙ 1 - 6 + 9 = 3 - 6 + 9 = 6

Answer 2

Y = 3x^2 + 6x + 9.
Since "A" is positive, the parabola opens upward and vertex is min. point
on the curve.
h = Xv = -B/2A = -6/(2*3) = -1.

K = Yv = Ymin. = 3x^2 + 6x + 9 = 3*(-1)^2 + 6*(-1) + 9 = 6.

V(h, k) = V(-1, 6).