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If the function represents a straight line and it’s known that and , find the value of .

- Jack
**Posts:**0**Joined:**Mon Jan 28, 2013 8:08 pm

The equation of a straight line is of the form

where is the slope and is the y-intercept.

The slope can be found by the following formula:

where are the points which the line passes through.

So, in your case the slope is

Therefore the function can be written as

Finally the function can be written as

So the value of is

Second Method

The GDC way using CASIO fx-9860 series is

MENU>STAT>Fill List1 with x-values (2 and -1) and List with the corresponding y-values (3 and 4)>CALC(F2)>REG(F3)>X(F1)> the output is

a=-0.3333333, b=3.666666.

Then you can graph the linear function y=-0.3333333x+3.666666

MENU>GRAPH>Y1=-0.3333333x+3.666666>DRAW(F6)>F5>Y-CAL>X=10 and the result is Y=0.333333

hope these help!!

where is the slope and is the y-intercept.

The slope can be found by the following formula:

where are the points which the line passes through.

So, in your case the slope is

Therefore the function can be written as

Finally the function can be written as

So the value of is

Second Method

The GDC way using CASIO fx-9860 series is

MENU>STAT>Fill List1 with x-values (2 and -1) and List with the corresponding y-values (3 and 4)>CALC(F2)>REG(F3)>X(F1)> the output is

a=-0.3333333, b=3.666666.

Then you can graph the linear function y=-0.3333333x+3.666666

MENU>GRAPH>Y1=-0.3333333x+3.666666>DRAW(F6)>F5>Y-CAL>X=10 and the result is Y=0.333333

hope these help!!

- miranda
**Posts:**268**Joined:**Mon Jan 28, 2013 8:03 pm

2 posts
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