1. ## [SOLVED] Graphing Porabolas

What am I doing wrong?

2. Originally Posted by minneola24
What am I doing wrong?
the table of values is wrong check your calculations again for the first and other values for the first: x^2-2x-1 (2)^2-2(2)-1=4-4-1=-1

might wanna complete the square to get the vertex:

x^2-2x-1
(x^2-2x)-1
(x^2-2x-1+1)-1
(x^2-2x)-1-1
(x-1)^2-2

3. Do you have a Ti? You could really limit calculation confusion by graphing it on there.
Here's how
y= -x^2-2x-1
Graph
2nd+Graph (Table)
Plot the points
Also. I suggest looking at the behavior of the graph depending on the first coefficient. If the beginning coefficient is negative, the graph is always facing downward. If the first coefficient is to the second degree (or an even degree for that matter) it's always going to be some sort of parabola. Also, any constants suggest a translation upward (positive constant) or downward (negative constant) of the graph from the parent function x^2.
Also keep in mind that parabolas (or any even powered function) are reflected across an axis of symmetry. In other words, if you take the fold-line as the x-value of the vertex (as achieved by -b/2a) it would be 1. Draw a dotted line at y=1 and fold the function over - it's exactly the same!
The standard form of a parabola is:
f(x)=ax^2+bx+c
f(x)=y; it's just function notation

So I hope maybe I helped you understand the nature of the graph more than just playing dot-to-dot. Keep both in mind in graphing! The nature can be explained simply by the calculations. But I think they are just good to remember in any case.

4. Originally Posted by minneola24
What am I doing wrong?
you applied the formula x = -b/(2a) incorrectly. note that a is negative