Given its parent function, f(x) = 2x, what is the equation of the function shown?
wow cool picture c: Hmm I'm not quite sure what the question is asking. Oh a graph, good.
haha thanks i think
Hmm I'm not really sure what the parent function thing is about... But it appears we have an exponential function graphed. Oh oh oh I see, you meant to write \(\large f(x)=2^x\) Ok ok ok now it makes sense. That is our parent function, and we have some alterations applied to it.
hhahah heheh
Normally, \(\large 2^x\) would have an `asymptote` at y=0. See the red dotted line? Not the one on your cheek, the one on the graph. Our asymptote appears to have been moved to y=2. So the function has been shifted `up` two units. We can write it like this,\[\large f(x)=2^x+2\] Ok let's see if there is anything else going on. :o hmm
If it was \(\large f(x)=2^x+2\), then it would pass through the point \(\large (0,\;3)\). But it doesn't. It's passing through \(\large (0,\;5)\). So we have some scalar being applied to it. \[\large f(x)=a2^x+2\] We can plug in our coordinate pair to solve for the missing value a. \[\large (0,\;5) \qquad \rightarrow \qquad 5=a \cdot 2^0+2\] What do you get for \(\large a\)? c:
2x+5?
It should be just a number c: We plugged in a value for our x, so there isn't an X when we're solving for a.
What is \(\large 2^0=\)
i got h(x) = 2^x +5 or 3(2)^x + 2 or 5(2)^x + 3 or 7(2)^x + 2
0
\(\large 2^0=1\) If you're unsure about that, throw it into your calculator c: So that gives us,\[\large 5=a +2\]Now solve for \(\large a\) ! :D
3
Yay gj, now plug that a value into the function we were working on. \(\large f(x)=a2^x+2\) And you should have your answer.
If this was confusing, there is another way to do it setting up a system of equations. But whatev :D
so 3(2)^x + 2?
yes
thank yewwwwwwww!!!
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