Recall what we know already about parabolas:
- They are the graphs of functions of the form $y = ax^{2} +bx + c$, where $a$, $b$, and $c$ are real numbers and $a \neq 0$.
- There is a distinguished point on the graph, called the vertex. It is either the lowest or highest point on the graph, depending on whether the graph opens upward or downward.
- The graph is symmetric across a vertical line through the vertex.
- The coefficient $a$ determines whether the graph opens upward or downward; upward if $a > 0$, downward if $a<0$.
- Otherwise $a$ influences the steepness or flatness of the graph. If $|a| > 1$ the graph is steeper than the standard parabola $y = x^2$; if $|a| < 1$ the graph is shallower and flatter than the standard graph.
- From shifting parabolas, we found that if $a=1$, the equation of the function has the form $y = (x-h)^2 + k$, where the vertex is $(h, k)$
- In general, the formula for any parabola can be put into the form $y = a(x-h)^2 + k$, where the vertex is the point $9h, k)$ and $a$ influences the steepness or shallowness of the graph and whether it opens upward or downward.
Today we take an equation in the form $y = ax^2 + bx +c$ and rewrite it into the “vertex form” $y = a(x-h)^2 + k$. I showed you two ways to do this. One is the way I was taught (and the way it is usually shown in textbooks), and the second way is the way the WeBWorK does it. You can use either method, whichever you prefer, but WeBWorK walks you through the second method.
Here is my slideshow with notes on the second method:
MAT1275parabolasCompleting Square-slideshow
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Finally, we saw a formula for finding the x-coordinate of the vertex, namely $x= -\frac{b}{2a}$ (which, I’m sure you will not be surprised to learn, comes from the Quadratic Formula).
Once we have found the x-coordinate, the y-coordinate of the vertex can be found by substituting into the formula for the parabola.
Example: for the parabola with formula $y = 3x^2 -4x +1$,
the x-coordinate of the vertex is given by $x = \frac{-(-4)}{2(3)} = \frac{4}{6} = \frac{2}{3}$
Find the y-coordinate: $y = 3\left(\frac{2}{3}\right)^2 – 4\left(\frac{2}{3}\right) + 1 = 3\left(\frac{4}{9}\right) -\frac{8}{3} + 1 = \frac{4}{3} – \frac{8}{3} + \frac{3}{3} = -\frac{1}{3}$
The vertex is the point $\left(\frac{2}{3}, -\frac{1}{3}\right)$
