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Question 902140: This question is involving Transforming Functions. I need help finding the domain and range's of the equation 2f(x-6)+1. Note: f(x) Domain: [-2, 1] Range: [-1, 3].
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! first find the equation for f(x)
we are given two points, (-2, -1) and (1, 3)
slope(change in y divided by change in x) is (3 - (-1)) / (1 - (-2) = 4/3
so far we have
y = 4x/3 + b
pick first point and solve for b
-1 = 4(-2)/3 +b
b = -3/3 + 8/3 = 5/3
equation is y = 4x/3 +5/3
use second point to check
3 = 4(1)/3 + 5/3
3 = 9/3 = 3
2f(x-6)+1 = 2(4(x-6)/3 +5/3) +1
2f(x-6)+1 = 2(4x-24)/3 +5/3) +1
2f(x-6)+1 = 8x-48/3 +10/3 +1
2f(x-6)+1 = (8x-48+13)/3
2f(x-6)+1 = 8x-35 / 3
we are given the domain for f(x) which is [-2, 1]
therefore the domain for 2f(x-6)+1 is [-8, -5] and range is [-33, -25]
2f(-8)+1 = -64-35/3 = -99/3 = -33
2f(-5)+1 = -40-35/3 = -75/3 = -25
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