Apparently, there is suppose to be 2 x-intercepts, which I really don't understand. How can the parabola cross the x axis twice when it has a y intercept of 0? Thanks to anyone who can help.
On
So the quadratic equation is: $$y \equiv f(x) = ax^2 + bx + c=0$$ The graph is: $$g(x) = (x,f(x))$$ If we know that we have a $y$ intecept of 0, then $g(0)=(0,0)$ lies on the parabola. Plugging it in: $$0=a0^2+b0+c \implies c=0 \implies ax^2+bx=0$$ Having $-1$ as an axis of symmetry means: $$a(-1+x)^2+b(-1+x)=a(-1-x)^2+b(-1-x)$$ $$ \implies a+ax^2-2ax-b+bx=a+ax^2+2ax-b-bx$$ $$\implies x(b-2a)=x(2a-b) \implies b=2a \implies f(x)=ax^2+2ax=ax(x+2)$$
For any $a \neq 0$
Perhaps this is what you want. Since the $y$ intercept is 0, one of the roots is $x=0$. Also, the axis of symmetry is $x=-1$ thus there is another root on the opposite side of the axis of symmetry. That root must be equally far from the first root which is $x=0$. Thus the root must be $x=-2$ So we get a quadratic: $$f(x)=x(x+2)$$