## 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 $\left( h,k \right)$. Therefore, the vertex of the parabola $f\left( x \right)={{\left( x-1 \right)}^{2}}+2$ is $\left( h,k \right)=\left( 1,2 \right)$. Step 2: The parabola is also symmetric with respect to the line $x=h$. So, the provided parabola is symmetric to $x=1$. Step 3: If $a>0$ , the parabola opens upward and if $a<0$ then the parabola opens downward. 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 upward as $a>0$. Step 4: So, we see that the above steps lead to the parabola that opens upwards and has a vertex at $\left( 1,2 \right)$. Therefore, the required parabola is shown above.