RE Tangent lines? ?Find an equation of the line tangent to the following curves at the given point. y= x +3;x = 3

Solution Step 1: 4x Given curve is y= x +3 ;x = 3 We have to find the equation of tangent line to the given curve at given point x=3 The equation of tangent line to the curve y=f(x) at given point (x ,y ) with slope m is given by 0 0 (y y ) = m(x x ) 0 0 Here we have given the point x =0 therefore the corresponding y is 0 0 y = 4(3) = 12 = 12 = 1 0 (3) +3 9+3 12 Therefore the point is (x ,y 0=(301) Step 2 Now to find slope dy = d ( 4x ) dx dx x +3 (x +3)dx4x)(4xdx(x +3) d u vdx(u)dx = 2 2 (dx v )= 2 ) 2 (x +3) (v) (x +3)(4)(4x)(2x) = (x +3) 2 2 = (4x +122x ) (x +3) (124x ) = (x +3)2