How Can I Solve for B in My Math Problem?

[Teh]

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.

 

[MarkFL]

a) We have:

$$y(2)=f(2)+g(2)=13+8=21$$

$$y'(2)=f'(2)+g'(2)=6+6=12$$

Using the point-slope formula:

$$y_{T}=12(x-2)+21=12x-24+21=12x-3$$

b) We have:

$$y(2)=f(2)-4g(2)=13-4(8)=-19$$

$$y'(2)=f'(2)-4g'(2)=6-4(6)=-18$$

Okay, you have the point $(2,-19)$ and the slope $-18$.

Using the point-slope formula, what do you get for the tangent line?

 

[Teh]

a) We have:

$$y(2)=f(2)+g(2)=13+8=21$$

$$y'(2)=f'(2)+g'(2)=6+6=12$$

Using the point-slope formula:

$$y_{T}=12(x-2)+21=12x-24+21=12x-3$$

b) We have:

$$y(2)=f(2)-4g(2)=13-4(8)=-19$$

$$y'(2)=f'(2)-4g'(2)=6-4(6)=-18$$

Okay, you have the point $(2,-19)$ and the slope $-18$.

Using the point-slope formula, what do you get for the tangent line?

will it be y = -18x - 55 ?

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will it be y = -18x - 55 ?

nvm caught my mistake y = -18x + 17

 

[MarkFL]

will it be y = -18x - 55 ?

No, I think you have made a very minor sign error...we have:

$$y_T=-18(x-2)-19$$

When you distribute the $-18$ to the $-2$, you get a positive value...;)

edit: Yes, now you have it. (Yes)

 

[Teh]

No, I think you have made a very minor sign error...we have:

$$y_T=-18(x-2)-19$$

When you distribute the $-18$ to the $-2$, you get a positive value...;)

edit: Yes, now you have it. (Yes)

for C) I got y = 24x + 4 ... make sure if i got it correct?

 

[MarkFL]

for C) I got y = 24x + 4 ... make sure if i got it correct?

c) We have:

$$y(2)=4f(2)=4(13)=52$$

$$y'(2)=4f'(2)=4(6)=24$$

Okay, you have the point $(2,52)$ and the slope $24$.

Using the point-slope formula, we find:

$$y_T=24(x-2)+52=24x+4\quad\checkmark$$

 

[Teh]

c) We have:

$$y(2)=4f(2)=4(13)=52$$

$$y'(2)=4f'(2)=4(6)=24$$

Okay, you have the point $(2,52)$ and the slope $24$.

Using the point-slope formula, we find:

$$y_T=24(x-2)+52=24x+4\quad\checkmark$$

thank you very much

 

[MarkFL]

The way I would have likely worked that as a student would have been to write:

$$h(x)=a\cdot f(x)+b\cdot g(x)$$

And so:

$$h(2)=a\cdot f(2)+b\cdot g(2)=13a+8b$$

$$h'(2)=a\cdot f'(2)+b\cdot g'(2)=6(a+b)$$

So, having the point $(2,13a+8b)$ and the slope $6(a+b)$, we have:

$$y_T=6(a+b)(x-2)+13a+8b=6(a+b)x+a-4b$$

Now, we have a formula to answer the 3 questions:

a) $a=b=1$

$$y_T=6(1+1)x+1-4=12x-3$$

b) $a=1,\,b=-4$

$$y_T=6(1-4)x+1+16=-18x+17$$

c) $a=4,\,b=0$

$$y_T=6(4+0)x+4-0=24x+4$$