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Simple Graph Problem.


Let's say that I have a linear equation, for example:

   y = 0.25x - 4

Now, at what value of 'x' should I start and end? Do I only need to do 2 points and just draw a long straight line throught both points (I'm aware that if this is the case, it wouldn't work for quadratic graphs)??

It's the same with Quadratic graphs, how to work out from an equation at what values of 'x' I need to start and end with?

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2 Solutions
Line are infinite.  You can plug in any value of x and get a value for y.  If you're trying to draw a graph, then you only need 2 points - choose values that are convenient to draw.  It's probably easiest if you choose values for x that are multiples of 4, since you'll get whole numbers for y.
Quadratics are also infinite.  However, the interesting bits are usually where the curve changes directions, so you'll need to find the minimum or maximum value.

If your quadratic is of the form
y = ax² + bx + c
Then we can take the the derivative, set it equal to 0, and find that the minimum or maximum has the x value
x = -b / (2a)

You'll also want to plug in a few values for x on either side of this number to define the shape of the curve.
You need 2 points and a ruler for a straight line (y = mx + c)

You need 3 points and a variable parabola tool (VPT) for plotting a quadratic (y = ax^2 + bx + c).
As VPTs are quite hard to come by, your better off drawing a lot of points and joining the dots.

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The "best" points to use for plotting a line are probably the x- and y-intercepts:

Set x = 0    ==>  y = 4

Set y = 0   ==>  0.25x - 4 = 0  ==>  x = 1

  Set y = 0   ==>  0.25x - 4 = 0  ==>  x = 16

For a quadratic (y = ax^2 + bx + c) you can work out where/if it crosses the x axis from
x= -b +/- sqrt(b^2 - 4ac)/2a

I think what you are asking might relate to exam-taking (judging from another post of yours).  I think you might be asking what to do, given your time constraints, if a question on an exam says, Plot y = 0.25x - 4, or Plot y = ax² + bx + c.

First, for the line:
As d-glitch said, the x and y-intercepts are the easiest points to use in general.  Label the points or the relevant cross-hatches on the axes.

In a rare case, the line will cross an axis with a large value and you won't want to draw so many cross-hatches.  Then choose some other convenient value.

As you said, sketch a line going through the two points.  Make sure it extends slightly beyond the points, and put arrows on both ends.

In exams, it's always a good idea to do some sort of independent check.  In this case, you could find a third point and make sure it's on the line you drew.  You could also calculate the slope and make sure that matches.

I don't know anything about the exam you will take, but often exams that cover this material present an equation for a line in several other forms, besides the form you mentioned in your question.  It would be good to be familiar with the other standard forms.  I hope you have a book with plenty of worked examples.

For the parabola:
As snoyes jw said, plot the extremum (i.e. the minimum or maximum).  Personally, what I would do next is see if the parabola is opening up or down.  There are a couple ways you could do this.  One is to choose a value of x near the extremum and see if it gives you a y-value that's bigger or smaller.

That's what I do when I'm plotting a parabola.  But I have a feeling they may want to see the focus and the directrix.  I have not done those in years.  I found a link that explains how:
(You can skip directly to "Graphing Method".)
However, you might be able to find some other site you like better.  Ideally, you should be working with a book that matches up pretty well with the exam you'll be taking.


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