To determine the horizontal intercepts, we'll set \(y=0\) and solve for \(x\text{:}\)

\begin{align*} 0\amp=-x^2+5x-7\\ \end{align*}

Noting that this equation cannot be solved using factoring, we'll use the quadratic formula:

\begin{align*} x\amp=\frac{-5\pm \sqrt{5^2-4(-1)(-7)}}{2(-1)}\\ x\amp=\frac{-5\pm \sqrt{25-28}}{-2}\\ x\amp=\frac{-5\pm \sqrt{-3}}{-2} \end{align*}

Since these numbers are not real, no solution exists to the equation and so there are no horizontal intercepts.

To determine the vertical intercept, we'll replace \(x\) with \(0\text{:}\)

\begin{align*} y\amp=-(0)^2+5(0)-7\\ y\amp=7 \end{align*}

Thus the \(y\)-intercept occurs at the point \((0,-7)\text{.}\)

The vertex will occur at \(x=-\frac{b}{2a}\text{:}\)

\begin{align*} x\amp=-\frac{5}{2(-1)}\\ \amp=\frac{5}{2} \end{align*}

To find the \(y\)-coordinate, we'll replace \(x\) with \(\frac{5}{2}\text{:}\)

\begin{align*} y\amp=-\left(\substitute{\frac{5}{2}}\right)^2+5\left(\substitute{\frac{5}{2}}\right)+7\\ \amp=-\frac{3}{4} \end{align*}

The vertex occurs at the point \(\left(\frac{5}{2},-\frac{3}{4}\right)\text{,}\) or at \((2.5,-0.75)\text{.}\)