Check Answer Question 15 Find \( b \) and \( c \) so that \( y=3 x^{2}+b x+c \) has vertex \( (-6,0) \). \( b=\square \) \( c=\square \)

To find \( b \) and \( c \) for the curve \( y = 3x^2 + bx + c \) to have the vertex at \( (-6,0) \), we can use the vertex formula for a parabola in the form \( y = ax^2 + bx + c \). The x-coordinate of the vertex is given by \( x = -\frac{b}{2a} \). Here, \( a = 3 \). Substituting the vertex point: \[ -6 = -\frac{b}{2(3)} \implies -6 = -\frac{b}{6} \implies b = 36. \] Now, to find \( c \), we substitute the vertex point \( (-6, 0) \) back into the equation: \[ 0 = 3(-6)^2 + 36(-6) + c. \] \[ 0 = 3(36) - 216 + c \implies 0 = 108 - 216 + c \implies c = 108. \] Thus, the values are: \( b=36 \) \( c=108 \)