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Teh
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View attachment 6125I was able to get A but, B got me good. How will I be doing this part because my professor have not shown us how to do this part.
will it be y = -18x - 55 ?MarkFL said:a) We have:
\(\displaystyle y(2)=f(2)+g(2)=13+8=21\)
\(\displaystyle y'(2)=f'(2)+g'(2)=6+6=12\)
Using the point-slope formula:
\(\displaystyle y_{T}=12(x-2)+21=12x-24+21=12x-3\)
b) We have:
\(\displaystyle y(2)=f(2)-4g(2)=13-4(8)=-19\)
\(\displaystyle y'(2)=f'(2)-4g'(2)=6-4(6)=-18\)
Okay, you have the point $(2,-19)$ and the slope $-18$.
Using the point-slope formula, what do you get for the tangent line?
nvm caught my mistake y = -18x + 17Teh said:will it be y = -18x - 55 ?
Teh said:will it be y = -18x - 55 ?
MarkFL said:No, I think you have made a very minor sign error...we have:
\(\displaystyle y_T=-18(x-2)-19\)
When you distribute the $-18$ to the $-2$, you get a positive value...;)
edit: Yes, now you have it. (Yes)
Teh said:for C) I got y = 24x + 4 ... make sure if i got it correct?
MarkFL said:c) We have:
\(\displaystyle y(2)=4f(2)=4(13)=52\)
\(\displaystyle y'(2)=4f'(2)=4(6)=24\)
Okay, you have the point $(2,52)$ and the slope $24$.
Using the point-slope formula, we find:
\(\displaystyle y_T=24(x-2)+52=24x+4\quad\checkmark\)
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