## Precalculus (6th Edition) Blitzer Use the steps shown below to determine the graph of the quadratic equation. Step 1: The quadratic function can be written as $f\left( x \right)=a{{\left( x-h \right)}^{2}}+k$ corresponding to which the graph is a parabola whose vertex is at $\left( h,k \right)$. Thus, the vertex of the parabola $f\left( x \right)=-{{\left( x-\frac{1}{2} \right)}^{2}}+\frac{5}{4}$ is $\left( h,k \right)=\left( \frac{1}{2},\frac{5}{4} \right)$. Step 2: The parabola is also symmetric with respect to the line $x=h$. So, the provided parabola is symmetric to $x=\frac{1}{2}$. Step 3: If $a>0$ , the parabola opens upwards and if $a<0$ then the parabola opens downwards. Also, if $\left| a \right|$ is small, the parabola opens more flatly than if $\left| a \right|$ is large. And, from the provided equation of the function, it is observed that the graph opens downwards as $a<0$. Step 4: So, we see that the above steps lead to the parabola that opens upwards and has a vertex at $\left( \frac{1}{2},\frac{5}{4} \right)$. Therefore, the required parabola is shown above.