## Lena772 2 years ago (a) find an equation of the tangent line to the graph of at the given point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of the graphing utility to confirm your results.

1. Lena772

function: y= cos 3x point: (pi/4, -sqrt2/2)

2. ganeshie8

where wer u stuck ?

3. ganeshie8

i saw u doing a similar problem the other day ? :)

4. Decart

take the derivative

5. Lena772

I just get confused when trig is added in

6. Decart

do you know what the derivative is

7. Lena772

dy/dx

8. Decart

-3sin3x

9. Decart

right

10. Lena772

yes

11. ganeshie8

so, slope = m = dy/dx = -3sin3x

12. ganeshie8

since u want the slope at point: (pi/4, -sqrt2/2) evaluate slope at x = pi/4

13. ganeshie8

dy/dx at x = pi/4 :- -3sin(3*pi/4) = ?

14. Lena772

-0.123

15. Lena772

@ganeshie8

16. ganeshie8

that doesnt look right you should get :- dy/dx at x = pi/4 :- -3sin(3*pi/4) = $$\large \frac{-3}{\sqrt{2}}$$

17. ganeshie8

now that u knw slope, m = $$\large \frac{-3}{\sqrt{2}}$$ point: (pi/4, -sqrt2/2) wats the equation of tangent in point slope form ?

18. Lena772

y-(-sqrt2/3)=(-3/sqrt2)(x-(pi/4)

19. Lena772

?

20. Lena772

@hartnn

21. ganeshie8

thats right !

22. Lena772

So that's for a? What would the graph look like for b? Do I just put that equation into Geogebra?

23. Lena772

@ganeshie8

24. ganeshie8

yup graph all 3 below :- 1) function: y= cos 3x 2) point: (pi/4, -sqrt2/2) 3) tangent : y-(-sqrt2/$$\color{red}{2}$$)=(-3/sqrt2)(x-(pi/4)

25. ganeshie8

u should get exactly like the one attached