Is it possible to graph linear functions like h(x) = 3x + 7 without doing a table of values and if no domain restricted is stated? If it is, how? And I know the slope is 7, which means the y-intercept is (0,7).

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- Dec 18th 2007, 09:51 AMMacleefMath - Functions
Is it possible to graph linear functions like h(x) = 3x + 7 without doing a table of values and if no domain restricted is stated? If it is, how? And I know the slope is 7, which means the y-intercept is (0,7).

- Dec 18th 2007, 09:58 AMbobak
If you want an accurate graph on graph paper,plot two points on the line using the formula a fair distance form each other and join them with a ruler.

- Dec 18th 2007, 10:18 AMxifentoozlerix
Actually, the slope here is 3 according to your equation. The y-intercept IS $\displaystyle (0,7)$ though as you stated. Whenever you have an equation of a line where x and y are dependent $\displaystyle (y=mx+b)$, you can find both an x-intercept $\displaystyle (x=-(b/m),y=0)$ and y-intercept $\displaystyle (x=0,y=b)$ which when connected and extended give the graph of the line.

- Dec 19th 2007, 12:14 AMAryth
With just the slope and the y-intercept, a graph is possible.

Most graphs that are in the form $\displaystyle y = mx + b$ have no restrictions which is why none are usually stated.

If you plot the slope and the y-intercept, a rough graph can be sketched if you use the slope.

The slope is:

$\displaystyle slope = \frac{rise}{run}$

If you plot the y-intercept and use the slope and go up the amount of the rise, which in the case is three, and then to the right or left by the amount of the run, you can find the next point. If we call the new x-value a, and the new y-value b, then we could use the following to obtain a new point (a,b) using:

$\displaystyle a = y-value + rise$

$\displaystyle b = x-value + run$

When you recalculate the above for every new point, you can graph the line without a data table.