SolutionThe vertical intercept is easiest choice to start with, since it is just the last term in the basic form: \(\left(0,-\frac{{9}}{{5}}\right)\). Next, let's consider the formula \(\left(\frac{-b}{{2a}},f\left(\frac{-b}{{2a}}\right)\right)\) to find the vertex:

\[ \solve{ \dfrac{-b}{{2a}} &=& \dfrac{\frac{{8}}{{5}}}{2\left(\frac{{1}}{{5}}\right)}\\ \dfrac{-b}{{2a}} &=&\dfrac{{8}}{{5}}\times \frac{{5}}{{2}}\\ \dfrac{-b}{{2a}} &=&4 } \]

Then plug this back into the function to get the vertical component:

\[ \solve{ f(4)&=&\dfrac{{1}}{{5}}(4)^2-\dfrac{{8}}{{5}}(4)-\dfrac{{9}}{{5}}\\ f(4)&=&\dfrac{{16}}{{5}}-\dfrac{{32}}{{5}}-\dfrac{{9}}{{5}}\\ f(4)&=&\dfrac{-25}{{5}}\\ f(4)&=&-5 } \]

Thus,the vertex is \((4,-5)\). Next, the horizontal intercepts. First I will set the function equal to zero and rearrange it to attempt factoring. If that fails, then I will use the quadratic formula:

\[ \solve{ 0 &=&\dfrac{{1}}{{5}}x^2-\dfrac{{8}}{{5}}x-\dfrac{{9}}{{5}}\\ 0&=&x^2-8x-9\\ 0&=&(x-9)(x+1)\\ \boxed{x=-1}&&\boxed{x=9} } \]

Thus, the horizontal intercepts are \((-1,0)\) and \((9,0)\). With all of this information, we can now sketch the graph: