graphing y = ax^2 + c
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Graphing y = ax2 + cBy L.D.
Before watching this,
watch my blog post on
Math Vocabulary of
Eighth Grade 2.
Problems
0 Problem 1: Graph y = x2
0 Problem 2: Graph y = 2x2
0 Problem 3: Graph y = ½x2
0 Problem 4: Graph y = -x2
0 Problem 5: Graph y = x2 - 40 Problem 6: Graph y = -x2 - 20 Problem 7: Graph y = 2x2 - 4
Problem 1
0 Graph y = x2
Problem 1
0 Graph y = x2
The first thing we need to do is to remember the x and y table we used to make for y = mx + b. We will do this on the below table always using the numbers -2, -1, 0, 1, 2.
X Y
-2-1 0 1 2
On the next page you will see how the table looks completed. Our problem is y = x2. We will do the problem by taking a number on the side of the x and place it in the x place of the problem. The answer of this will go in the y problem place.
Problem 1
0 Graph y = x2
X Y
-2-1 0 1 2
Now we need to graph the problem. We will make our marking points using the x in the x axis and the y in the y axis. The graph will be on the next page.
41 0 1 4
Now we need to find the axis of symmetry (my
graph is not perfect, so we must imagine it is
completely equal in its line lengths) and the
vertex.
We must remember that the axis of symmetry is the line
that splits the graph completely in half,
imagining my graph was perfect, our axis of
symmetry would be directly in the middle. From now on
this presentation I shall make the axis of symmetry
purple.
The vertex is green. This is the vertex because it is the
lowest part of the graph that does not go onward. If the parabola was flipped then the vertex would be
the highest part because it wouldn’t have lines going onward. From now on this presentation I shall make
the vertex green.
Problem 2
0 Graph y = 2x2
Problem 2
0 Graph y = 2x2
Now we must use our table again and get points which we will plot.
X Y
-2-1 0 1 2
Problem 2
0 Graph y = 2x2
Now we must graph this and find the vertex and axis of symmetry. X Y
-2-1 0 1 2
81 0 18
Problem 3
0 Graph y = ½x2
Problem 3
0 Graph y = ½x2
We will again use the table and then graph. We will find the vertex and axis of symmetry again.
X Y
-2-1 0 1 2
X Y
-2-1 0 1 2
4½ 0½4
Mini LessonThere is no such thing as negative 0. If
you were to have a problem like –(0)2. The answer
would not be -0, but only 0 as zero trumps all.
Problem 4
0 Graph y = -x2
Problem 4
0 Graph y = -x2
Now we need to make a table again, and then find the vertex and line of symmetry.
X Y
-2-1 0 1 2
When you replace the x and put a number in its place we will make the problem (using r as an example) look like –(r)2. We will remember PEMDAS as we do this.
Problem 4
0 Graph y = -x2
Now we need to make a table again, and then find the vertex and line of symmetry.
X Y
-2-1 0 1 2
X Y
-2-1 0 1 2
X Y
-2-1 0 1 2
-(-2)2
-(-1)2
-(0)2
-(1)2
-(2)2
-4-1 0 -1 -4
The vertex is in the highest position as it
is the place where lines do not go
onward.
Problem 5
0 Graph y = x2 – 4
Problem 5
0 Graph y = x2 – 4
Now we need to make a table again, and then find the vertex and line of symmetry.
X Y
-2-1 0 1 2
X Y
-2-1 0 1 2
0-3-4-30
Problem 6
0 Graph y = -x2 + 2
Problem 6
0 Graph y = -x2 + 2
X Y
-2-1 0 1 2
-2121-2
X Y
-2-1 0 1 2
Problem 7
0 Graph y = 2x2 - 4
Problem 7
0 Graph y = 2x2 – 4
Now we need to make a table again, and then find the vertex and line of symmetry.
X Y
-2-1 0 1 2
X Y
-2-1 0 1 2
4-2 -4-2 4
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