A) Find the slope and equation of the tangent line to the graph of the equation at a given point.

1. y=2x2+4x at point (-2,0)

2. y=2x3+3 at point(1,5)

3. y=9—Jr2 at x=2 17%|

ACTIVITIES

A) Find the derivative of the given function at a given number

1. y=3x2-2 at f'(1)

2. y=x2—l at f'(2)

3. f (x)= % , solve for f'(—2)

4. y =2x2—2x, solve for f'(0)

A.

1. Slope = -4 

Tangent y = -4x - 8 

2. Slope = 6

Tangent y = 6x - 1  

3. Slope = -4 

Tangent y = -4x + 13 

A 

1. f^'(1) = 6 

2. f'(2) = 4 

3. f'(-2) = 2 

4. f'(0) = -2 

Step-by-step explanation

A O y = 2 x + 4 2 at paint (- 2, 0) = (20, , y, ) Differentiating, with respect to x = 40 +4 ( at ( - 2,0) = 4 x (- 2 ) + 4 = - 8+4 = -4 Slope Vat (- 2, 0) = -4 Answer Equation of the tangent Line J - YI = yat (-20) ( X - 26, ) J - 0 = - 4 ( 20 - (-2) ) y = - 4 ( 2( +2 ) 1 = - 4x - 8 Equation of the tangent line 4 = - 4x - 8 Ansley

3. y = 9- 2 at DC = 2 Differentiating with respect to y' = - 2 20 when 2 = 2 , y = 9- 22 = 9-4 = 5 when x = 2 , y = 5 Point ( 2 , 5 ) = (26 , , 41 ) Jat ( 2,5 ) = - 2x2 = - 4 slope yat ( 2, 5 ) = - 4 Equation of tangent line y - y, = Jat ( 2 , 5) ( 2( - 20 , ) y - 5 = -4 (x-2) y - 5 = - 42( + 8 1 = - 40C+ 8+5 1 = - 42 +13 Equation of tangent line y= - 4x+ 13

A O y = 3x -2 at P'(l) 4' = 6x - 0 y = 620 f (I ) = 6 x 1 = 6 P ( 1 ) = 6 Ansley 2 at P ' ( 2) 9 = 20- 0 1 =
20C P ( 2 ) = 2 X 2 = 4 f ( 2 ) = 4 3 ) p(x ) = P ( - 2 ) = ? DC + 1 P ( x ) = (x + 1 ) dx d (2x ) 20 9 ( 2(+ 1
) (x(+ 1 )2 = (2+ 1 )2 - 20 x (1+0 ) 20C+ 2- 2 0 = (2+ 1)2 (2(+ 1)2

P ( x ) = 2 (2( + 1)2 f'( - 2 ) = 2 2 ( - 2+ 1)2 ( - 1)2 2 = 2 f ' ( - 2 ) = 2 Angley 4 y = 2 02- 20C y' = 4x - 2
P ( 0 ) = 4 x0 - 2 0 - 2 - 2 P' ( 0 ) = - 2 Angley

Please see attached file