# If I am given a regression equation (with no data) how do I show the scatter plot graph on a TI-83?  The actual question is match the regression equation y=0.00115x+2.93 with the graph.  I know which way it slopes but I don't understand where it crosses the y-intercept as my pictures of graphs don't make it clear. You should rememeber that a graph intercepts y axis at x = 0, hence, you need to substitute 0 for x in the given equation of the function such that:

`y=0.00115*0+2.93 => y = 2.93`

Hence, the graph intercepts y axis at `(0,2.93).`

You also should find the x intercept...

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You should rememeber that a graph intercepts y axis at x = 0, hence, you need to substitute 0 for x in the given equation of the function such that:

`y=0.00115*0+2.93 => y = 2.93`

Hence, the graph intercepts y axis at `(0,2.93).`

You also should find the x intercept substituting 0 for y in equation of the function such that:

`0=0.00115x+2.93 => 0.00115x = -2.93 => x =-2.93/0.00115`

`x = -2547.8260`

Hence, the graph intercepts x axis at `(-2547.8260,0).`

Hence, evaluating y and x intercepts yields (0,2.93) and (-2547.8260,0).

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