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f(x) = 5x + 1 f(1) = 5(1) + 1 ... replace every x with 1 f(1) = ???
close but no
"5(1) + 1" means "5 times 1 plus 1"
yeah f(1) = 6
"f−1(3) when f(x) = 2x+3" is a fancy way of saying "plug in f(x) = 3 and solve for x" f(x) = 2x+3 3 = 2x+3 3-3 = 2x+3-3 ... Subtract 3 from both sides. 0 = 2x x = ???
yep x = 0 leads to f(x) = 3
solve 3y − 7 = y + 5 for y. Tell me what you get
y = 6 is correct
from part a) we got f(1) = 6
ok so 1 and 3 :)
I have one last one and I'm done for tonight!
show me what you have so far
Ok. so I guess a positive end behavior for f(x)+2
how do you know it's positive?
i just assumbed bc the leading co. is positive
we don't know anything about the original function f(x). So we can't say if it has positive or negative end behavior
what we can say is that the end behavior won't change if we add 2 to f(x)
true. I agree
f(x)+2 just shifts f(x) up 2 units
on the other hand (-1/2)*f(x) flips f(x) over the x axis and compresses it vertically by a factor of 2
so (-1/2)*f(x) will have its end behavior flipped
what can you say about the y-intercept?
It would be (0,2)
but do we know what the y-intercept of f(x) is?
we don't know what the y-intercept of f(x) is
but whatever it is, it is shifted up 2 units for f(x)+2
yes. And what about the -1/2 one?
so a wide parabola?
it gets wider when you compress it vertically, yes
But I don't understand how it is increasing and the regions where it does part>>
f(x) isn't given, so we cannot find the increasing/decreasing intervals. Whatever they are, they don't change when going to f(x)+2. Everything shifts up which is why the intervals don't change
with (-1/2)*f(x), the intervals swap. Whatever was decreasing is now increasing and vice versa.
Oh:O !! That makes sooo much more sense! Thanks:)
is that it? We are done with the question
yeah I think so