###### Answer9.3.22.1

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 $$(2.5,-0.75)\text{.}$$

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