Answer:
First, let's find the equations for our lines:
A linear relationship can be written as:
y = a*x + b
where a is the slope and b is the y-axis intercept.
For a line that passes through the points (x1, y1) and (x2, y2), the slope can be written as:
a = (y2 - y1)/(x2 - x1).
f(x) passes through (5,0) and (10, 10), then the slope is:
a = (10 - 0)/(10 - 5) = 10/5 = 2.
then we have:
y = 2*x + b
And when x = 5, we have y = 0.
0 = 2*5 + b
0 = 10 + b
b = -10
Then the equation for f(x) is:
y = f(x) = 2*x - 10.
Now for g(x) we have the points:
(3, 0) and (2, 10)
a = (10 - 0)/(2 - 3) = -10
y = -10*x + b
0 = -10*3 + b
b = 30.
y = g(x) = -10*x + 30.
A) Ok, the transformations:
Transformation 1 or T1.
f(x) = 2*x - 10
g(x) = -10*x + 30.
Then, we start with f(x):
First, we can move f(x) up 4 units and get:
f'(x) = 2*X - 6
Now we can dilate f(x) with a scale factor of -5 from the origin, now we get:
f''(x) = -5*f'(x) = -10*x + 30.
And this is g(x).
Transformation 2 or T2.
Move f(x) up 10 units, so now we have:
f'(x) = 2*x
Do a reflection over the x-axis, so the sign of y changes, and now we get:
f''(x) = -2*x
Do a dilation of scale factor 5
f'''(x) = 5*-2*x = -10*x
Now do a vertical translation of 30 units up.
f''''(x) = -10*x + 30 = g(x).
These are two transformations that start with f(x) and end with g(x).
B) Ok, as i was writting the transformations i already solved them, so this part is already done.
C) the equation for the transformations are:
T1) g(x) = -5*(f(x) + 4)
T2) g(x) = -(f(x) + 10)*5 + 30
