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I'm not sure how the transformation works in this problem. $$f(x) = 3(1+e^x)-2$$ I thought it was a vertical stretch by 3 and vertical translation 1 down (adding 1 and 2?? idk)

So how does the transformation work here? (assuming that $$f(x)=e^x$$ is the original function)

QuantumPi
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  • $e^x$ is your "base" function. $1+e^x$ shifts it up by $1$. $3(1+e^x)$ well, stretch what you have vertically by a factor of $3$. and then $3(1+e^x)-2$ will shift it down by $2$. Alternatively $3(1+e^x)-2=3e^x + 1$ which would be the same as stretching it by a factor of three and then shifting it up by $1$. (Those aren't contradictory: when we stretch $(1+e^x)\to 3(1+e^x)$ we are stretching the initial shift of $1$ to now stretch to a shift of $3$. Shifting down by $2$ will result in a net shift (after stretch) of $1$. – fleablood Feb 02 '22 at 21:19
  • For peace of mind: consider then points: $(-big, e^{-big}=verysmall>0),(-2,e^{-2}\approx \frac 17),(-1,e^{-1}\approx \frac 13),(0,e^0=1), (1,e),(2,e^2\approx 7),(big,e^{big}=verybig)$. These are $(x,e^x)$. Vertical shift them $(x,1+e^x)$ so $(-big, 1+verrysmall), (-2,\approx 1\frac 17),(0,2),(\approx1,3.7),(big, verybig +1)$. Stretch them by $3$: $ (-big,3+3verysmall),(-2, \approx 3\frac 37),(0,6),(1,\approx 11.1),(big, 3verybig + 3)$. Now vertical shift down: $(-big, 1+3verysmall),(-2,\approx 1\frac 37),(0,4),(2,\approx 9.1=3\times 2.7+1),(big,3verybig + 1)$. – fleablood Feb 02 '22 at 21:35

1 Answers1

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Your task is very simple. First, you can write: $$f(x)=3e^x+1$$ So, you can follow two steps. Firstly, stretch by $3$. Then, traslate by $v=(0,1)$.

Note that if you traslate by $v$ and then strecltch, you are going to obtain a different result.

Matteo
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