To find the vertex of the quadratic function [tex]\( y = 3x^2 + 12x + 7 \)[/tex], we need to use the vertex formula for a parabola given by [tex]\( y = ax^2 + bx + c \)[/tex]. The x-coordinate of the vertex can be found using the formula:
[tex]\[ h = -\frac{b}{2a} \][/tex]
Here, [tex]\( a = 3 \)[/tex] and [tex]\( b = 12 \)[/tex]. Plugging in these values:
[tex]\[ h = -\frac{12}{2 \times 3} \][/tex]
[tex]\[ h = -\frac{12}{6} \][/tex]
[tex]\[ h = -2 \][/tex]
Next, we need to find the y-coordinate of the vertex. This can be done by substituting [tex]\( h = -2 \)[/tex] back into the original equation [tex]\( y = 3x^2 + 12x + 7 \)[/tex]. So, let's compute [tex]\( y \)[/tex] when [tex]\( x = -2 \)[/tex]:
[tex]\[ y = 3(-2)^2 + 12(-2) + 7 \][/tex]
[tex]\[ y = 3(4) + 12(-2) + 7 \][/tex]
[tex]\[ y = 12 - 24 + 7 \][/tex]
[tex]\[ y = -12 + 7 \][/tex]
[tex]\[ y = -5 \][/tex]
Therefore, the y-coordinate [tex]\( k \)[/tex] is [tex]\( -5 \)[/tex].
So, the vertex of the quadratic function [tex]\( y = 3x^2 + 12x + 7 \)[/tex] is:
[tex]\[ (-2, -5) \][/tex]
Hence, the vertex form of the quadratic function is:
[tex]\[ y = 3(x + 2)^2 - 5 \][/tex]
