Question

( y=(x+1) quad(x-3) )
When it meets x aris ( rightarrow quad y=0 )
( 0=(x+1)(x-3) )
( x=-1 quad, quad x=3 )
gradient at ( x=-1, frac{d y}{d x}=(x+1)(1)+(x-3)(1) )
( frac{d y}{d x} mid=frac{2 x-2 mid}{x=-1}=[-4] )
gradient at ( x=3 ), ( mid 2 x 3-2=4 )
( tan ^{-1}left(frac{8}{15}right)=theta )

# The angle between the tangents at those points on the curve y=(x + 1)(x-3) where it meets x-axis, is

Solution