#### Question

Consider the following.

f (2/3)=-15/2, f (-4)=-14

(a) Write the linear function f such that it has the indicated function values.

f(x)=_______________

and how can I graph the result

This is another way of asking for the equation of a line through 2 points. The points are (2/3,-15/2) and (-4,-14).

So first find the slope. Slope = (y2-y1)/(x2-x1). In this case, slope=(-14-(-15/2))/(-4-2/3)

=(-13/2)/(-14/3)=39/28

so, f(x)=(39/28)x+b

Substitute second point and solve for b

-14=(39/28)(-4)+b

-14=(-39/7)+b

-14+39/7=b

b=-98/7+39/7=-59/7=-8 3/7

Check:

-14=(39/28)(-4)-59/7

-14=-156/28 -236/28=-392/28=-14 It checks!

-15/2=(39/28)(2/3)-59/7

-15/2=26/28-236/28=-15/2 It checks!

Graph it as a line with slope (39/28) [about 1.4] and a y-intercept of -8 3/7.

#1

Let y = f(x)

x1 = 2/3, y1 = -15/2

x2 = -4, y2 = -14

m = (y2 - y1)/(x2 - x1) = (-14 - (-15/2))/(-4 - 2/3)

m = (-13/2)/(-14/3)

m = 13/2 * 3/14

m = 39/28

y - y1 = m(x - x1)

y - (-15/2) = 39/28 (x - 2/3)

y + 15/2 = (39/28)x - 13/14

y = (39/28)x - 13/14 - 15/2

y = (39/28)x - 59/7

f(x) = (39/28)x - 59/7

Ans: (39/28)x - 59/7

To graph take any two points which satisfy the equation. For example, plot points (2/3, -15/2) and (-4, -14)

Plot these two points and join them by a straight line.

#2

Let f(x)= mx + c

When x = 2/3

-15/2 = 2m/3 + c (1)

When x = -4

-14 = -4m+ c (2)

(1) -(2) gives

-15/2 - -14 = 2m/3 - - 4m

-15/2 + 28/2 = 2m/3 + 12m/3

13/2 = 14m/3

m = 3/14 * 13/2 = 39/28

Sub. m = 39/28 & x = -4 into (2)

-14 = -4 * 39/28 + c

c = -14 + 39/7 = -98/7 + 39/7 = -59/7

Therefore f(x) = 39x/8 - 59/7

#3