When we multiply two functions, the degree is the sum of the original degrees. So, since the degree of the product is 2, we have two cases:
- One of the function has already degree 2, and the other is constant (degree 0)
- Both functions are linear.
The first case is actually impossible, because otherwise the sum would have degree 2 as well. So, we know that both
and
are linear. In other words, we have
![f(x)=ax+b,\quad g(x)=cx+d](https://img.qammunity.org/2021/formulas/mathematics/middle-school/ht3k87fbkdmmf5fbftyfxudh3my38ipe7h.png)
for some
![a,b,c,d \in \mathbb{R}](https://img.qammunity.org/2021/formulas/mathematics/middle-school/xl2ho9m5r31tw93g0bugciovluwh58x8p6.png)
We know that the sum is
![h(x)=4x=f(x)+g(x)=(a+c)x+(b+d)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/orro9lr6fyrk4vwaqr0hsenfdhqngtvozx.png)
we deduce that
![a+c=4,\quad b+d=0](https://img.qammunity.org/2021/formulas/mathematics/middle-school/kd8k0wxn368df77g4qs7kwe894vhmvrziy.png)
So, we know that:
- Both functions must be quadratic. FALSE, otherwise the product would have degree 4;
- Both functions must have a constant rate of change. TRUE, linear functions have a constant rate of change;
- Both functions must have a y-intercept of 0. FALSE, it is only required that the sum of the y-intercepts is 0, they don't have to be both zero;
- The rate of change of f(x) and g(x) must be opposites. FALSE, their sum must be 4;
- The y-intercepts of f(x) and g(x) must be opposites. TRUE, their sum must be zero.