Suppose a curve has y axis coordinates as $(0,0)$ and x axis coordinates as $(15/2,0)$. How do I find the equation of the graph using either $y=k(x-a)(x-b)$ or $y=k(x-a)^2+b$.
I have used this formula, $y=k(x-a)(x-b)$ and at one point I get $k=0$ for which I failed to get any equation. I have no idea how to go further down(needless to say I just have started learning about curves). Anyways, the equation is $y=15x-2x^2$.
EDIT: I just figured, the two points where the curve meets the coordinate axes are $x=0$ and $x$=$ \frac{15}{2} $ so we have two factors here; $(x-0)(15-2x)$ so if we multiply the two factors we obtain the final equation. I would like to know what are the other ways to get equations.
We will use $y=k(x-a)(x-b).$
Clearly $k\neq0,$ since otherwise will have $y=0$.
Then the point $(0,0)$ gives $0=k(0-a)(0-b)=kab$. So either $a=0$ or $b=0$ or both.
Suppose that $a=0$ (or similarly $b=0$), then we will have $y=kx(x-b)$ and then the point $\color{red}{(\frac{15}{2},0)}$ (and not $(0,\frac{15}{2})$ like above) gives $0=k\frac{15}{2}(\frac{15}{2}-b).$ Then since $k\neq0$ we must have $b=\frac{15}{2}.$
Thus we have $y=kx(x-\frac{15}{2})=\frac{kx}{2}(2x-15)$ for any $k\neq 0$. So choosing $k=-2$, we get $y=-x(2x-15)=15x-2x^2$ for example.
Note that since the roots of the quadratic are $x=0$ and $x=\frac{15}{2}$ then it is of the form $y=k(x-0)(x-\frac{15}{2})=\frac{kx}{2}(2x-15)$ for some $k\neq 0$.