Find the coordinates of the vertex for the parabola defined by the given quadratic function.
-$$f ( x ) = 3 x ^ { 2 } - 6 x - 5$$
A) $( 1 , - 8 )$
B) $( - 1,4 )$
C) $( - 2,19 )$
D) $( 2,1 )$

Question

Quizplus · Accepted Answer

The vertex of a parabola defined by $f(x) = ax^2 + bx + c$ can be found using the formula $(- \frac{b}{2a}, f(- \frac{b}{2a}))$. For the given function $f(x) = 3x^2 - 6x - 5$, $a = 3$ and $b = -6$. Thus, the x-coordinate of the vertex is $- \frac{-6}{2*3} = 1$. Substituting $x = 1$ into the original equation gives $f(1) = 3(1)^2 - 6(1) - 5 = 3 - 6 - 5 = -8$. Therefore, the vertex is $(1, -8)$.

Find the Coordinates of the Vertex for the Parabola Defined f(x)=3x2−6x−5f ( x ) = 3 x ^ { 2 } - 6 x - 5f(x)=3x2−6x−5

Find the Coordinates of the Vertex for the Parabola Defined $f ( x ) = 3 x ^ { 2 } - 6 x - 5$