College Algebra, Chapter 2, 2.3, Section 2.3, Problem 14

Determine an appropriate viewing rectangle for the equation $y = 0.3x^2 + 1.7x -3$ and use it to draw the graph.

We can determine the appropriate viewing rectangle easier by getting the $x$ and $y$-intercept at the equation. So, if we set $y=0$

$\begin{equation} \begin{aligned} 0 &= 3x^2 + 1.7x - 3 && \text{Divide both sides by } 0.3\\ \\ 0 &= x^2 + \frac{17}{3} x - 10 && \text{Solve for } x\\ \\ x &= \frac{-17 + \sqrt{649}}{6} \text{ and } x = \frac{-17-\sqrt{649}}{6} \end{aligned} \end{equation}$

Thus, the $x$-intercepts are at $\displaystyle \left( \frac{-17+\sqrt{649}}{6}, 0 \right)$ and $\displaystyle \left( \frac{-17-\sqrt{649}}{6},0 \right)$
Or $(1.4126,0)$ and $(-7.0792,0)$

Next, solving for $y$-intercept, where $x = 0$

$\begin{equation} \begin{aligned} y &= 0.3(0)^2 + 1.7(0) -3\\ \\ y &= -3 \end{aligned} \end{equation}$


The $y$-intercept is $(0,-3)$
Thus, we set the appropriate viewing rectangle to be $[-10,5]$ by $[-6,10]$