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: Identify the parabola's direction

The function \( f(x) = \frac{1}{2}x^2 + 5x + 6 \) is a quadratic function. The coefficient of \( x^2 \) is \( \frac{1}{2} \), which is positive. So, the parabola opens upward.

Step2: Find the vertex's x - coordinate

For a quadratic function in the form \( f(x)=ax^2 + bx + c \), the x - coordinate of the vertex is given by \( x=-\frac{b}{2a} \). Here, \( a = \frac{1}{2} \) and \( b = 5 \).
\[
x=-\frac{5}{2\times\frac{1}{2}}=-\frac{5}{1}=- 5
\]

Step3: Determine the increasing interval

For a parabola that opens upward, the function is increasing to the right of the vertex (i.e., for \( x \) values greater than the x - coordinate of the vertex). Since the vertex has an x - coordinate of \( x=-5 \), the function is increasing on the interval \( (-5,\infty) \).

Answer:

B. \((-5, \infty)\)