questions 050907

Question:When eliminating the variable 'C' in a system of equations does it absolutely necessary to subtract equation 1 from equation 2 & equation 1 from equation three? Would the answer be incorrect if one were to subtract equation 2 from equation 3 in place of the original second step?" "Ex
1) 80=4a+16b+c
2) 40=3a+9b+c
3) 30=2a+4b+c
40=3a+9b+c
- 80=4a+16b+c
= -40=-1-7
30=2a+4b+c
- 80=4a+16b+c
= -50=-2-12b+c

As long as you don't use the same combination of equations (e.g., subtract equation 1 from equation 2 then subtract equation 2 from equation 1) you will get the same solution no matter which two pairs you choose.  If for example you subtracted equation 2 from equation 3, and subtracted equation 1 from equation 3, the solution would work out just fine.


Does the quadratic formula work if we only have two data points and just plug in one as zero or x?

The quadratic formula is what we use to find solutions to an equation of the form a t^2 + b t + c = 0.

I'm going to assume that your question addresses whether we can obtain a quadratic model with just two data points.

Since a quadratic model is of the form y = a t^2 + b t + c, with three parameters a, b and c, we need at least three equations in order to solve for a, b and c and get our model.
So two data points would not be sufficient.

If we happen to know that (0, 0) is a data point, of course, then we could indeed plug in 0 for x and y. We would end up with 0 = c, so we would conclude that our parameter c must be 0.

However unless we have information that tells us that (0, 0) is a data point, we can't use it as such.

My question is, on problem #2 of the assigned worksheet you only want us to graph each equation once, Right? The second (last) paragraph for this problem is just clarification, I hope, for the preceeding statement.

Good question. Do try to include a copy of the problem, so when I post my answer it will be clear exactly which problem you are asking about. Another reason is so I'll know exactly what you are asking about.

The problem I think you are asking about asks the following:

Determine the location of the vertex of each of the following quadratic functions:

y = x^2 + x + 1

y = x^2 + 2x + 1

y = x^2 + 3x + 1

y = x^2 + 4x + 1

y = -2 x^2 + 4x + 3

Sketch a graph for each function showing 3 points: the vertex and the points 1 unit to the right and 1 unit to the left of the vertex.

Sketch the graph of each function, based only on the 3 points you have sketched and your knowledge that the graph is a parabola.

To clarify:

You are asked to first sketch only 3 points for each function.
You are then asked to sketch the graph based on just these 3 points.

One of the questions for homework says: Sketch the graph that would be formed if your graph is y=x^2 was stretched vertically by a factor of 2, etc... are you just making the graph y=x^2 wider and skinnier?

If a graph is stretched vertically by factor 2, that means every point of the graph moves twice as far from the horizontal axis.

The points (-1, 1), (0, 0) and (1, 1) are the basic points of the y = x^2 graph.

If this graph is vertically stretched by factor 2, then the points become (-1, 2), (0, 0) and ( 1, 2).

This has the effect of making the graph appear skinnier (but not wider, which would be contradict 'skinnier').

I understand that the packet of paper you gave us is to study and learn but do we have to do work in it to turn in or do we do it on the sids just to understand it better. I was unclear on what this is used for (homework or study materials)

The instructions I believe I gave were to read the material and work the exercises. That will be the standard assignment with the packets.