List the transformations of the function j(x) = 3(-7(x+2))-4
1 answer:
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Answer:
Intercepts:
x = 0, y = 0
x = 1.77, y = 0
x = 2.51, y = 0
Critical points:
x = 1.25, y = 4
x = 2.17 , y = -4
x = 2.8, y = 4
Inflection points:
x = 0.81, y = 2.44
x = 1.81, y = -0.54
x = 2.52, y = 0.27
Step-by-step explanation:
We can find the intercept by setting f(x) = 0
where n = 0, 1, 2,3, 4, 5,...
Since we are restricting x between 0 and 3 we can stop at n = 2
So the function f(x) intercepts at y = 0 and x:
x = 0
x = 1.77
x = 2.51
The critical points occur at the first derivative = 0
or
where n = 0, 1, 2, 3
Since we are restricting x between 0 and 3 we can stop at n = 2
So our critical points are at
x = 1.25,
x = 2.17
,
x = 2.8,
For the inflection point, we can take the 2nd derivative and set it to 0
We can solve this numerically to get the inflection points are at
x = 0.81,
x = 1.81,
x = 2.52,