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# What is the equation for the graph of g(x)g left p

## Tags: Mathematik

20:59 Uhr, 09.01.2022

Tag!
I need you point me in the right direction here. where to focus my attention first. thank you in advance for your help!

The graph of the equation $f\phantom{\rule{0.12em}{0ex}}\left(x\phantom{\rule{0.12em}{0ex}}\right)=2x\phantom{\rule{0.12em}{0ex}}$ is reflected over the $x\phantom{\rule{0.12em}{0ex}}-a\phantom{\rule{0.12em}{0ex}}\xi \phantom{\rule{0.12em}{0ex}}s\phantom{\rule{0.12em}{0ex}}$ and translated three units to the right to become g(x)gas depicted below.

What is the equation for the graph of $g\phantom{\rule{0.12em}{0ex}}\left(x\phantom{\rule{0.12em}{0ex}}\right)g\phantom{\rule{0.12em}{0ex}}$

Für alle, die mir helfen möchten (automatisch von OnlineMathe generiert):
"Ich möchte die Lösung in Zusammenarbeit mit anderen erstellen."

23:25 Uhr, 09.01.2022

Hi,

to reflect $\left(x\phantom{\rule{0.12em}{0ex}}\right)=2x\phantom{\rule{0.12em}{0ex}}$ on the x-axis you just multiply $f\phantom{\rule{0.12em}{0ex}}\left(x\phantom{\rule{0.12em}{0ex}}\right)$ by $\left(-1\right)$. Thus the reflected function is ${f\phantom{\rule{0.12em}{0ex}}}_{r\phantom{\rule{0.12em}{0ex}}}\left(x\phantom{\rule{0.12em}{0ex}}\right)=-2x\phantom{\rule{0.12em}{0ex}}$

To shift ${f\phantom{\rule{0.12em}{0ex}}}_{r\phantom{\rule{0.12em}{0ex}}}\left(x\phantom{\rule{0.12em}{0ex}}\right)$ to the right you subtract $3$ from the $x\phantom{\rule{0.12em}{0ex}}$-variable: $g\phantom{\rule{0.12em}{0ex}}\left(x\phantom{\rule{0.12em}{0ex}}\right)=-2\left(x\phantom{\rule{0.12em}{0ex}}-3\right)=-2x\phantom{\rule{0.12em}{0ex}}+6$

greetings,
pivot

07:33 Uhr, 10.01.2022

www.wolframalpha.com/input/?i=plot+-2x+%2C+-2%28x-3%29

15:37 Uhr, 13.01.2022

thank you both. I have not forgotten about the question. $i\phantom{\rule{0.12em}{0ex}}$ am just studying the topic. it is kind of new to me. I study math on my own and with textbooks. Don't go to school. I work. So I am sort of pressed for time. Thank you for the tips. I am studying everything about graph transformation and all that. I will get back if in doubt.

02:33 Uhr, 16.01.2022

Tag, if $i\phantom{\rule{0.12em}{0ex}}$ didn't reply was because $i\phantom{\rule{0.12em}{0ex}}$ was studying the topic. it is a new one for me
$i\phantom{\rule{0.12em}{0ex}}$ input all the equations given in the answer choice and was clear for me to see that choice $A\phantom{\rule{0.12em}{0ex}}\left($ the first one) was the right equation.

Now, I would like to know how to do it without the graphing calculator. pivot, can you explain the reasoning behind what you wrote. I do not understand it well. thank you so much.

18:02 Uhr, 17.01.2022

Let´s say the linear function is $f\phantom{\rule{0.12em}{0ex}}\left(x\phantom{\rule{0.12em}{0ex}}\right)=m\phantom{\rule{0.12em}{0ex}}x\phantom{\rule{0.12em}{0ex}}+b\phantom{\rule{0.12em}{0ex}}$. Now you reflect it at the x-axis. For this purpose you just multiply the function by $-1$: ${f\phantom{\rule{0.12em}{0ex}}}_{r\phantom{\rule{0.12em}{0ex}}}\left(x\phantom{\rule{0.12em}{0ex}}\right)=-m\phantom{\rule{0.12em}{0ex}}x\phantom{\rule{0.12em}{0ex}}-b\phantom{\rule{0.12em}{0ex}}$.
$\cdot \phantom{\rule{0.167em}{0ex}}$If then the function is translated $d\phantom{\rule{0.12em}{0ex}}$ units to the right you subtract $d\phantom{\rule{0.12em}{0ex}}$ from $x\phantom{\rule{0.12em}{0ex}}$ : $g\phantom{\rule{0.12em}{0ex}}\left(x\phantom{\rule{0.12em}{0ex}}\right)=-m\phantom{\rule{0.12em}{0ex}}\left(x\phantom{\rule{0.12em}{0ex}}-d\phantom{\rule{0.12em}{0ex}}\right)-b\phantom{\rule{0.12em}{0ex}}=-m\phantom{\rule{0.12em}{0ex}}x\phantom{\rule{0.12em}{0ex}}+m\phantom{\rule{0.12em}{0ex}}d\phantom{\rule{0.12em}{0ex}}-b\phantom{\rule{0.12em}{0ex}}$
$\cdot \phantom{\rule{0.167em}{0ex}}$If then the function is translated $d\phantom{\rule{0.12em}{0ex}}$ units to the left you add $d\phantom{\rule{0.12em}{0ex}}$ to $x\phantom{\rule{0.12em}{0ex}}$: $g\phantom{\rule{0.12em}{0ex}}\left(x\phantom{\rule{0.12em}{0ex}}\right)=-m\phantom{\rule{0.12em}{0ex}}\left(x\phantom{\rule{0.12em}{0ex}}+d\phantom{\rule{0.12em}{0ex}}\right)-b\phantom{\rule{0.12em}{0ex}}=-m\phantom{\rule{0.12em}{0ex}}x\phantom{\rule{0.12em}{0ex}}-m\phantom{\rule{0.12em}{0ex}}d\phantom{\rule{0.12em}{0ex}}-b\phantom{\rule{0.12em}{0ex}}$

numerical example: $f\phantom{\rule{0.12em}{0ex}}\left(x\phantom{\rule{0.12em}{0ex}}\right)=3x\phantom{\rule{0.12em}{0ex}}+5$
Reflecting f(x) at the x-axis and then translate the reflected function two units to the left.

${f\phantom{\rule{0.12em}{0ex}}}_{r\phantom{\rule{0.12em}{0ex}}}\left(x\phantom{\rule{0.12em}{0ex}}\right)=-3x\phantom{\rule{0.12em}{0ex}}-5$, $g\phantom{\rule{0.12em}{0ex}}\left(x\phantom{\rule{0.12em}{0ex}}\right)=-3\left(x\phantom{\rule{0.12em}{0ex}}+2\right)-5=-3x\phantom{\rule{0.12em}{0ex}}-11$

15:34 Uhr, 22.01.2022

thank you so much. Ich verstehe jetzt. I have also been watching some tutorials and videos about graphing transformation of exponential equations.

15:49 Uhr, 22.01.2022

Freut mich, dass das klar ist. Transformationen von Exponentialfunktionen sind natürlich was anderes. If something unclear about that issue, feel free to ask a new question.