View Full Version : sat functions

Brynlee

08-22-2013, 07:09 PM

I need help with the following question,please.

If the function f(x) represents a straight line and it’s known that f(2)=3 and f(-1)=4, find the value of f(10) .

Thanks

The equation of a straight line is of the form

f(x)=ax+b

where a is the slope and b is the y-intercept.

The slope can be found by the following formula:

a=\frac{y_{2}- y_{1}}{ x_{2}- x_{1}} where [tex]( x_{1}, y_{1}), ( x_{2}, y_{2}) are the points which the line passes through.

So, in your case the slope is

a=\frac{f(-1)- f(2)}{-1-2}=\frac{4- 3}{-1-2}=\frac{1}{-3}=-\frac{1}{3}

Therefore the function can be written as

f(x)= -\frac{1}{3} x+b

f(2)=3 \Rightarrow -2\frac{1}{3} +b =3 \Rightarrow b =3+\frac{2}{3}=\frac{11}{3}

Finally the function can be written as

f(x)= -\frac{1}{3} x+\frac{11}{3}

So the value of f(10) is

f(10)= -10\frac{1}{3} +\frac{11}{3}=\frac{1}{3}

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

Powered by vBulletin® Version 4.2.2 Copyright © 2019 vBulletin Solutions, Inc. All rights reserved.