If you know very little about the topic, a good way would be to simply make a table of values for x and y
pick reasonable small values for x and evaluate to get the y
e.g.
x y
0 -3
1 -4
2 -3 ------> 2^2 - 2(2) - 3 = 4-4-3 = -3
3 0
4 5 ----> 4^2-2(4)-3 = 16-8-3 = 5
5 12
-1 0
-2 5
-3 12 ---> (-3)^2 - 2(-3)-3 = 9+6-3 = 12
now you have 9 points, plot them and you should see the outline of the parabola. Join the points with a smooth curve, don't join them with straight lines.
Notice that (1 , -4) is the vertex of the parabola
your graph should look like this
http://www.wolframalpha.com/input/?i=plot+y+%3D+x%5E2+-+2x+-+3
Please, can anyone explain the concept of graphing quadratic functions to me?
For example, how must I graph gradratic functions such as y = x^2 - 2x - 3?
If anyone could also include step-by-step solving methods, I would truly appreciate it.
2 answers
Reiny's method is fine, but the "standard" method is like this:
Find the vertex, which is the highest or lowest point of the parabola. Use -b/2a first.
y = ax^2 + bx + c
y = x^2 - 2x - 3
2/2 = 1, axis of symmetry.
y = 1^2 - 2(1) - 3
y = 1 - 2 - 3
y = -4
Vertex: (1, -4)
y-intercept: y = 0^2 - 2(0) - 3
y-intercept: -3
Choose another number that is on the same side as the y-intercept and vertex.
y = 4^2 - 2(4) - 3
y = 16 - 8 - 3
y = 5
(4, 5) is another point.
Graph using the points (4,5) and (1,4) and the y-intercept -3.
Find the vertex, which is the highest or lowest point of the parabola. Use -b/2a first.
y = ax^2 + bx + c
y = x^2 - 2x - 3
2/2 = 1, axis of symmetry.
y = 1^2 - 2(1) - 3
y = 1 - 2 - 3
y = -4
Vertex: (1, -4)
y-intercept: y = 0^2 - 2(0) - 3
y-intercept: -3
Choose another number that is on the same side as the y-intercept and vertex.
y = 4^2 - 2(4) - 3
y = 16 - 8 - 3
y = 5
(4, 5) is another point.
Graph using the points (4,5) and (1,4) and the y-intercept -3.