Question

How do I graph 7/2x-2?

Answer 1

First, I'm not sure if this is supposed to be read as
`y = 7/2x - 2`
or
`y = 7/(2x) - 2`

So we will plot both.

THE LINE
The function is in slope intercept form. We know that at `x=0`, `y=-2`. We also know that the line goes up 7 and over 2, so we also know that it runs into the point (2, 5). We can take these lines, connect them and get our line.

THE INVERSE FUNCTION
We will plot this by transforming it relative 1/x. We know what 1/x looks like. We can think: what does 1/(2x) look like?

For 1/x to get to 1/2, we need to plug in two but in order to get to 1/2 with 1/(2x), we only need to plug in one, so it is shrinking twice as fast! However, we also have the 7 up top which slows down the shrinking by the same logic and makes it ultimately be slower falling than 1/x.

Then we have the -2, which shifts the whole function downward by 2.

Answer 2

Assumption: you mean: `y=7/2 x-2`
see explanation.

Explanation:

NOT USING SHORTCUT METHODS

This is an equation of a straight line

`color(blue)("The line crosses the x-axis (intercept) when "y" is 0")`

Set: `0=7/2 x-2`

If you multiply 0 by 2 you still end up with 0. Lets 'get rid' of the 2 from `7/2 x`. Multiply everything in both sides of the equals by 2

`0=7x-4`

Add 4 to both sides

`4=7x+0`

Divide both sides by 7

`4/7=x color(white)("ddddd")larr color(white)("d")x_("intercept")`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("The line crosses the y-axis (intercept) when "x" is 0")`

`y=7/2 x-2`

Set `x=0` giving:

`y=(7/2 xx0) -2`

`y=-2color(white)("ddddd")larr color(white)("d")y_("intercept")`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
DRAW A STRAIGHT LINE THROUGH THOSE POINTS.
Make sure you label your points and put a title on the graph.

Example: Graph of `y=7/2x-2`

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