A guest Post by Paul Pierce and Ashley Ross

With the advances in calculator technology, some developmental and college-level math courses are restricting the use of any type of graphing or programmable calculators. This is to help students avoid becoming dependent on their calculators for both simple arithmetic and graphing. So, some teachers are going “old school” and forbidding the use of calculators in the classroom. Therefore, it is imperative that students learn efficient methods for finding important values, as well as graphing functions, without the help of their calculator. One type of function that appears in many courses is the quadratic function, and one of the most critical points on the graph of a quadratic function is the vertex.

## Fundamental Concepts of the Graph of a Quadratic Function

For the function $f(x)=ax^2+bx+c$ with $a\not=0$, the graph is a smooth, continuous curve called a parabola. This parabola opens upward if $a > 0$ or opens downward if $a < 0$. The vertex $(h,k)$ of the graph is the only turning point on the parabola, which makes it a critical point. The $y$-coordinate $k$ of the vertex represents the minimum value of the function if $a>0$, or the maximum value of the function if $a<0$.

The point $(h,k)$ may be found using the formulas $h=\frac{-b}{2a}$ and $k=\frac{bh}{2}+c$, which begin to show the importance of the vertex. We give two examples:

**Example 1.** For $y=x^2+6x+3$, find the vertex $(h,k)$.

First find $h$ using $h=\frac{-b}{2a}=\frac{-6}{2(1)}=-3$.

Next find $k$ using $k=\frac{bh}{2}+c=\frac{(6)(-3)}{2}+3=-9+3=-6$.

So, the coordinates of the vertex of the parabola are $(-3, -6)$. Observe from the graph that this vertex is the lowest point on the parabola, which means that $k = -6$ is the minimum value of the function.

**Example 2.**For $y=-2x^2+8x-5$, find the vertex $(h,k)$.

First find h using $h=\frac{-b}{2a}=\frac{-8}{2(-2)}=2$.

Next find k using $k=\frac{bh}{2}+c=\frac{(8)(2)}{2}-5=8-5=3$.

So, the coordinates of the vertex of the parabola are $(2, 3)$. Note that this vertex is the highest point on the graph, which illustrates that $k = 3$ is the maximum value of this function.

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