Answer
$(3, 10)$ is the maximum point.
Work Step by Step
First, we find the standard form. To find the standard form of some function $f(x) = ax^2+bx+c$, it would be $f(x) = a(x-h)^2+k$ where $h = -\frac{b}{2a}$ and $k = f(h)$. This is a standard result derived in the book simplifies the algebra and gives a closed form for the end result.
For this problem, $a= -\frac{1}{3},b = 2, c = 7$. Plugging above, we get $h = 3, k = f(3) = 10$ and hence $f(x) = -\frac{1}{3}(x-3)^2+10.$
The vertex can be easily deduced from the standard form; it is the point $(h, k)$ so in this case $(3, 10)$.
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 $(3, 10)$ is a maximum.
