I'm a math teacher and was creating an exam for my Algebra 1 students when I tried picking an equation that had integer Intercepts (both x- and y-), as well as extrema. I wanted to do so because I figured integer values would be most easily identifiable and comfortable for my students.
But that got me wondering... How would one guarantee that a given quadratic in standard form meets those three constraints?
I've got some ideas, but curious to know of other approaches that might be out there.
Let $f(x)=a(x-r)(x-s)$, where $a,r,s$ are integers and $a\ne 0$.
The vertex has $x$-coordinate ${\Large{\frac{r+s}{2}}}\!$,$\;$so the only additional condition you need is for $r+s$ to be even.
For example, using $a,r,s=1,3,-5$, you get $$f(x)=1(x-3)(x+5)=x^2+2x-15$$ which has $x$-intercepts $(3,0),(-5,0)$,$\;y$-intecept $(0,-15)$, and vertex $(-1,-16)$.