What is the minimum value of $y$ if $y=3x^2+6x+9?$
First, complete square as follows: $$y=3x^2+6x+9=3\left(x^2+2x\right)+9.$$ To complete the square, we need to add $\left(\frac{2}{2}\right)^2=1$ after the $2x.$ So we have $$y+3=3\left(x^2+2x+1\right)+9.$$ This gives $$y=3\left(x+1\right)^2+6.$$ Now, since $\left(x+1\right)^2\ge0,$ the minimum value is when the squared term is equal to $0.$ So the minimum value is $$y=3\left(x+1\right)^2+6=3\cdot0+6=\boxed{6}.$$
6
Level 4
Algebra
Problem: What is the minimum value of $y$ if $y=3x^2+6x+9?$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the minimum value of the quadratic function \( y = 3x^2 + 6x + 9 \), we recognize that this is a standard form of a quadratic equation \( ax^2 + bx + c \), where \( a = 3 \), \( b = 6 \), and \( c = 9 \). Since \( a > 0 \), the parabola opens upwards, indicating that there is a minimum point.

The x-coordinate of the vertex of a parabola given by \( ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \). Substituting the values of \( a \) and \( b \):

1. Calculate \( -\frac{b}{2a} \):
   - \( -\frac{6}{2 \cdot 3} = -\frac{6}{6} = -1 \).

Next, we substitute \( x = -1 \) back into the original equation to find the corresponding \( y \)-value:

2. Calculate \( y \) at \( x = -1 \):
   - \( y = 3(-1)^2 + 6(-1) + 9 \).
   - \( y = 3(1) - 6 + 9 \).
   - \( y = 3 - 6 + 9 = 6 \).

Thus, the minimum value of \( y \) occurs at \( x = -1 \) and is equal to \( 6 \). This confirms that the final answer is \( 6 \).<|END_OF_LATENT|> The final answer is $6$.