Solution This function is in the basic form, so the "easy" to get info is the vertical intercept: \( \left(0,-\frac{{6}}{{5}}\right)\). To ge the horizontal intercepts and vertex will require a bit more work. Let's start with the vertex, which can be found using the \(\left(\frac{-b}{{2a}},f\left(\frac{-b}{{2a}}\right)\right)\) formula:

\[ \solve{ \dfrac{-b}{{2a}} &=&\dfrac{-\frac{{22}}{{5}}}{2\left(-\frac{{7}}{{10}}\right)}\\ \dfrac{-b}{{2a}}&=&-\dfrac{{22}}{{5}}\times -\dfrac{{5}}{{7}}\\ \dfrac{-b}{{2a}}&=&\dfrac{{22}}{{7}} } \] To get the other coordinate, we plug this into the equation: \[ \solve{ f\left(\dfrac{{22}}{{7}}\right) &=&-\dfrac{{7}}{{10}}\left(\dfrac{{22}}{{7}}\right)^2+\dfrac{{22}}{{5}}\times\dfrac{{22}}{{7}}-\dfrac{{6}}{{5}}\\ &=&-\dfrac{{7}}{{10}}\times\dfrac{22^2}{7^2}+\dfrac{22^2}{5\times 7} -\dfrac{{6}}{{5}}\\ &=&-\dfrac{22^2}{10\times7}+\dfrac{22^2}{5\times 7} -\dfrac{{6}}{{5}}\\ &=&\dfrac{-22^2+2\times 22^2-6\times 14}{10\times 7}\\ &=&\dfrac{{400}}{{70}}\\ &=&\dfrac{{40}}{{7}}\\ } \] Thus the vertex is located at \(\left(\dfrac{{22}}{{7}},\dfrac{{40}}{{7}}\right)\).

Next, we lok for the horizontal intercepts by setting the function equal to 0. While we could technically use the quadratic formula right away, I will take a few steps to eliminate the fractions and see if the result is factorable. If not, then I will use the quadratic formula.

\[ \solve{ 0&=& -\dfrac{{7}}{{10}}x^2+\dfrac{{22}}{{5}}x-\dfrac{{6}}{{5}}\\ 0&=&-7x^2+44x-12\\ 0&=&7x^2-44x+12 } \] This is not easily factorable (if at all) so I will apply the quadratic formula to this equation instead. \[ \solve{ x&=&\dfrac{44\pm\sqrt{44^2-4(7)(12)}}{{14}}\\ x&=&\dfrac{44\pm\sqrt{1936-336}}{{14}}\\ x&=&\dfrac{44\pm\sqrt{{1600}}}{{14}}\\ x&=&\dfrac{44\pm40}{{14}}\\ x=\dfrac{{84}}{{14}}&&x=\dfrac{{4}}{{14}}\\ \boxed{x=6}&&\boxed{x=\dfrac{{2}}{{7}}} } \] Thus, the horizontal intercepts are \(\left(\frac{{2}}{{7}},0\right)\) and \((6,0)\).

With this information, we can now sketch the graph. One final piece of information we can take away is that because the leading coefficient is negative, the parabola should open down. We note this as one means of checking our work.