A quadratic function has zeros at x = 4 and x = -3. What is the x-coordinate of its vertex?

A parabola is symmetric about its axis, so the vertex lies midway between the zeros. The $x$-coordinate of the vertex is $\frac{4+(-3)}{2}=\frac{1}{2}=0.5$, choice F. Check with the equation $f(x)=(x-4)(x+3)=x^2-x-12$: the vertex is at $x=-\frac{b}{2a}=\frac{1}{2}$. Choice G ($1$) is the sum of the zeros, which skips the division by $2$. Choice H ($3.5$) averages the absolute values, $\frac{4+3}{2}$. Choice J ($-3.5$) is the same value with the sign changed. The vertex must lie between $-3$ and $4$.