## Precalculus (6th Edition) Blitzer So, for the provided quadratic equation $f\left( x \right)=2{{x}^{2}}+4x-3$ , when compared with the standard form of the quadratic equation $f\left( x \right)=a{{x}^{2}}+bx+c$ , we get: \begin{align} & a=2, \\ & b=4, \\ & c=-3 \end{align} 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{4}{2\times 2} \\ & =-1 \end{align} Or the y-coordinate can be calculated as: \begin{align} & y=2{{\left( -1 \right)}^{2}}+4\left( -1 \right)-3 \\ & =2-4-3 \\ & =-5 \end{align} So, the vertex is $\left( -1,-5 \right)$. Step 3: And the above steps lead to the parabola that opens upwards and has a vertex at $\left( -1,-5 \right)$ and it intersects x-axis at $\left( -1+\sqrt{\frac{5}{2}} \right)$ and $\left( -1+\sqrt{\frac{5}{2}} \right)$ , and y-axis at -3. Thus, the required parabola is as shown above.