Pre-calc/trig
1 answer:
You might be interested in
Answer with explanation:
The given function is:
y=x³+3 x²-x-3
=x²(x+3) -1(x+3)
= (x²-1)(x+3)
= (x-1)(x+1)(x+3)
Put, y=0 , gives
⇒ (x-1)(x+1)(x+3)=0
≡→x-1=0 ∧ → x+1=0 ∧→x+3=0
≡x=1, -1, -3
So, there are three breaking points in the curve.
1.⇒Putting, x=1 , in the function
y(1)=1³+3×1²-1 -3
=1 +3-4
=0
2. ⇒Putting, x=-1 , in the function
y(-1)=(-1)³+3×(-1)²-(-1) -3
= -1 +3+1-3
=0
3.⇒Putting, x=-3 , in the function
y(-3)=(-3)³+3×(-3)²-(-3) -3
=-27+27+3-3
=0
So, These are points where curve crosses the X axis, which are , (1,0),(-1,0) and (-3,0).
But turning points are those points where curve takes the turn , that is, if it is increasing function then it starts decreasing or becomes constant function after that point, and if it is a decreasing function then it becomes increasing or constant function.
So,turning points of the given function by looking at the graph of the function by approximation
Option B:→(-2,3) and (0, -3)
