Answer:
Yes, I agree
Step-by-step explanation:
For the cubic function
A cubic function is represented as:
![f(x)=ax^3+bx^2+cx+d](https://tex.z-dn.net/?f=f%28x%29%3Dax%5E3%2Bbx%5E2%2Bcx%2Bd)
A cubic function may have 1, 2 or 3 x intercepts. This is shown below
For 3 x intercepts
Equate y to 0
![x^3 - x = 0](https://tex.z-dn.net/?f=x%5E3%20-%20x%20%3D%200)
Expand
![x(x^2 - 1) = 0](https://tex.z-dn.net/?f=x%28x%5E2%20-%201%29%20%3D%200)
Express
as difference of two squares
![x(x - 1)(x +1 ) = 0](https://tex.z-dn.net/?f=x%28x%20-%201%29%28x%20%2B1%20%29%20%3D%200)
<em>x = 0 or 1 or -1</em>
For 2 x intercepts
![y = x^3 - x](https://tex.z-dn.net/?f=y%20%3D%20x%5E3%20-%20x)
Equate y to 0
![(x-5)^2(x+7) = 0](https://tex.z-dn.net/?f=%28x-5%29%5E2%28x%2B7%29%20%3D%200)
Expand
![(x-5)(x-5)(x+7) = 0](https://tex.z-dn.net/?f=%28x-5%29%28x-5%29%28x%2B7%29%20%3D%200)
<em>x= 5 or x = -7</em>
For 1 x intercept
![y = x^3](https://tex.z-dn.net/?f=y%20%3D%20x%5E3)
Equate y to 0
![x^3 = 0](https://tex.z-dn.net/?f=x%5E3%20%3D%200)
Take cube roots of both sides
![x = 0](https://tex.z-dn.net/?f=x%20%3D%200)
It has been shown above that a cubic function may have 1, 2 or 3.
So, I agree to the statement
For the quadratic function
A quadratic function will not have any x intercept when the function can not be factorized;
E.g.
![y = x^2 + x + 17](https://tex.z-dn.net/?f=y%20%3D%20x%5E2%20%2B%20x%20%2B%2017)
<em>The above function has no x intercept.</em>
A quadratic function will have at least 1 x intercept when the function can be factorized;
E.g.
![y = x^2- 6x + 9](https://tex.z-dn.net/?f=y%20%3D%20x%5E2-%206x%20%2B%209)
Equate y to 0
![x^2- 6x + 9 = 0](https://tex.z-dn.net/?f=x%5E2-%206x%20%2B%209%20%3D%200)
Expand
![x^2 - 3x - 3x + 9 = 0](https://tex.z-dn.net/?f=x%5E2%20-%203x%20-%203x%20%2B%209%20%3D%200)
![(x - 3)(x-3) = 0](https://tex.z-dn.net/?f=%28x%20-%203%29%28x-3%29%20%3D%200)
![x = 3](https://tex.z-dn.net/?f=x%20%3D%203)
We've shown that a quadratic may have no x intercept, and it may also have x intercept(s)
Hence, I agree to both statement