The graph of the parabola is shown below:
To find the vertex we need to complete the squares of the equation, let's do this:
[tex]\begin{gathered} y=15x^2-x-6 \\ y=15(x^2-\frac{x}{15})-6 \\ y=15(x^2-\frac{x}{15}+(-\frac{1}{30})^2)-6-15(\frac{1}{30})^2 \\ y=15(x-\frac{1}{30})^2-6-\frac{1}{60} \\ y=15(x-\frac{1}{30})^2-\frac{361}{60} \end{gathered}[/tex]
Therefore, the vertex of the parabola is at (1/30,-361/60) which can be express as (0.033,-6.017).
The y intercept happens when x=0, plugging this in the equation we have:
[tex]\begin{gathered} y=15(0^2)-0-6 \\ y=-6 \end{gathered}[/tex]
then, the y-intercept have coordinates (0,6).
Finally, the x-intercepts happens when y=0, plugging this in the equation and solving for x we have:
[tex]\begin{gathered} 15x^2-x-6=0 \\ \text{ Using the quadratic formula we have:} \\ x=\frac{-(-1)\pm\sqrt{(-1)^2-(4)(15)(-6)}}{2(15)} \\ x=\frac{1\pm\sqrt{361}}{30} \\ x=\frac{1\pm19}{30} \\ \text{ then} \\ x=\frac{1+19}{30}=\frac{20}{30}=\frac{2}{3} \\ \text{ or} \\ x=\frac{1-19}{30}=-\frac{18}{30}=-\frac{3}{5} \end{gathered}[/tex]
Therefore, the x-intercepts of the parabola are (2/3,0) and (-3/5,0) that can be written as (0.667,0) and (-0.6,0). Finally the parabola opens up, as we see on the graph.
