Find the vertex and axis of symmetry of the graph of the function.
-$$f ( x ) = x ^ { 2 } + 3 x + 6$$
A) $\left( \frac { 3 } { 2 } , \frac { 51 } { 4 } \right) ; x = \frac { 3 } { 2 }$
B) $( - 1,6 ) ; x = - 1$
C) $( 3,24 ) ; x = 3$
D) $\left( - \frac { 3 } { 2 } , \frac { 15 } { 4 } \right) ; x = - \frac { 3 } { 2 }$

Question

Quizplus · Accepted Answer

The vertex form of a quadratic function is $f(x) = a(x - h)^2 + k$, where $(h, k)$ is the vertex. The axis of symmetry is $x = h$. For the given function $f(x) = x^2 + 3x + 6$, we can find the vertex by completing the square or using the formula $h = -\frac{b}{2a}$ and $k = f(h)$, where $a = 1$, $b = 3$, and $c = 6$. Thus, $h = -\frac{3}{2(1)} = -\frac{3}{2}$ and $k = (-\frac{3}{2})^2 + 3(-\frac{3}{2}) + 6 = \frac{9}{4} - \frac{9}{2} + 6 = \frac{15}{4}$. Therefore, the vertex is $\left(-\frac{3}{2}, \frac{15}{4}\right)$ and the axis of symmetry is $x = -\frac{3}{2}$.

Find the Vertex and Axis of Symmetry of the Graph f(x)=x2+3x+6f ( x ) = x ^ { 2 } + 3 x + 6f(x)=x2+3x+6

Find the Vertex and Axis of Symmetry of the Graph $f ( x ) = x ^ { 2 } + 3 x + 6$