Question

over which interval is the graph of $f(x) = \frac{1}{2}x^2 + 5x + 6$ increasing? \\(\bigcirc\\ (-6.5, \infty)\\) \\(\bigcirc\\ (-5, \infty)\\) \\(\bigcirc\\ (-infty, -5)\\) \\(\bigcirc\\ (-infty, -6.5)\\)

Explanation:

Step1: Recall the vertex form of a parabola

For a quadratic function \( f(x) = ax^2 + bx + c \), the x - coordinate of the vertex is given by \( x=-\frac{b}{2a} \). For the function \( f(x)=\frac{1}{2}x^{2}+5x + 6 \), we have \( a=\frac{1}{2} \) and \( b = 5 \).

Step2: Calculate the x - coordinate of the vertex

Using the formula \( x=-\frac{b}{2a} \), substitute \( a=\frac{1}{2} \) and \( b = 5 \):
\( x=-\frac{5}{2\times\frac{1}{2}}=-\frac{5}{1}=- 5 \)

Step3: Determine the direction of the parabola and the increasing interval

Since \( a=\frac{1}{2}>0 \), the parabola opens upwards. A parabola that opens upwards is increasing to the right of its vertex (i.e., for \( x>-5 \) or the interval \( (-5,\infty) \)). We can also check from the graph: the vertex is at \( (-5,-6.5) \), and as we move to the right of \( x = - 5 \), the y - values of the function increase.

Answer:

\( (-5,\infty) \) (corresponding to the option: \( \boldsymbol{(-5,\infty)} \))