## Precalculus (6th Edition) Blitzer

We have the given quadratic equation is $f\left( x \right)=2{{x}^{2}}-7x-4$. Step 1: Determine how the parabola opens: Note that a, the coefficient of ${{x}^{2}}$ , is 2. 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: evaluate the vertex The x-coordinate can be calculated as: \begin{align} & x=-\frac{b}{2a} \\ & =-\frac{-7}{2\times 2} \\ & =\frac{7}{4} \end{align} And the y-coordinate can be calculated as: \begin{align} & y=2{{\left( \frac{7}{4} \right)}^{2}}-7\left( \frac{7}{4} \right)-4 \\ & =\frac{49}{8}-\frac{49}{4}-4 \\ & =-\frac{81}{8} \end{align} Thus, the vertex is $\left( \frac{7}{4},-\frac{81}{8} \right)$. Step 3: Above steps lead to the parabola that opens upwards and has a vertex at $\left( \frac{7}{4},-\frac{81}{8} \right)$ and it intersects the x-axis at $4$ and $-\frac{1}{2}$ , and the y-axis at -4. Thus, the required parabola is as shown above.