Answer
$(-2, -8)$ is the minimum point.
Work Step by Step
First, we find the standard form. To find the standard form of some function $h(x) = ax^2+bx+c$, it would be $h(x) = a(x-p)^2+k$ where $p = -\frac{b}{2a}$ and $k = h(p)$. This is a standard result derived in the book simplifies the algebra and gives a closed form for the end result.
For this problem, $a= \frac{1}{2},b = 2, c = -6$. Plugging above, we get $p = -2, k = h(-2) = -8$ and hence $h(x) = \frac{1}{2}(x+2)^2-8.$
The vertex can be easily deduced from the standard form; it is the point $(p, k)$ so in this case $(-2, -8)$.
The maximum or minimum of a quadratic function is attained at the vertex. To determine whether the vertex is a maximum or minimum, we look at $a$:
- if $a>0$, this tells us the function will grow towards positive infinity and hence, the vertex is a minimum.
- if $a<0$, this tells us that the function will grow towards negative infinity and hence, the vertex is a maximum.
Looking at $a$ in this case tells us that the vertex $(-2, -8)$ is a minimum.
