## Precalculus (6th Edition) Blitzer For the provided quadratic equation $f\left( x \right)={{x}^{2}}+6x+3$ , when compared with the standard form of the quadratic equation $f\left( x \right)=a{{x}^{2}}+bx+c$ , we get: \begin{align} & a=1, \\ & b=6, \\ & c=3 \end{align} And use the steps shown below to determine the graph of the quadratic equation. Step 1: Determine how the parabola opens: Note that a, the coefficient of ${{x}^{2}}$ , is 1. 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. Now, from the provided equation of the function, it is observed that the graph opens upwards as $a>0$. Step 2: Calculate the vertex: The x-coordinate can be calculated as: \begin{align} & x=-\frac{b}{2a} \\ & =-\frac{6}{2\times 1} \\ & =-3 \end{align} Or, the y-coordinate can be calculated as: \begin{align} & y={{\left( -3 \right)}^{2}}+6\left( -3 \right)+3 \\ & =9-18+3 \\ & =-6 \end{align} Therefore, the vertex is $\left( -3,-6 \right)$ Step 3: So, the above steps lead to the parabola that opens upwards and has a vertex at $\left( -3,-6 \right)$ and it intersects x-axis at $\left( -3+\sqrt{6} \right)$ and $\left( -3-\sqrt{6} \right)$ , and y-axis at 3. Therefore, the required parabola is as shown above.