Sketch the Graph of $-x^3+27$ 
Not sure if this is the right way to solve. 
The question ask to sketch the graph of each function by transforming the graph of an appropriate function of the form $y=x^n$. Indicate all $x$- and $y$- intercepts on each graph.
Is this right?  
 A: *

*$-x^3$ is the reflection about y-axis of $x^3$

*$-x^3+27$ is just a transformation in positive y direction 

*to get the x-intercept we put $y=0$ and then $$-x^3+27=0\implies -(x-3)(x^2+3x+9)=0\implies x=3$$ 



A: To get a more accurate graph, you'll also want to find the $x$-intercept(x): where your function value $f(x) = 0$. $$27-x^3 = 0 \iff x^3 - 27 = 0 \iff (x-3)(x^2 + 3x + 9) = 0$$
This will happen at $x = 3$. ($x^2 + 3x + 9$ has no real roots, so there is one and only one real-valued x-intercept.) So you can graph the point $(3, 0)$ on your graph, as well.
*Note: your factoring is a little bit off. 
A: Semsem has provided the graphs; here is how you would find the intercepts.
For the $y$ intercept, this is just when $x$ is evaluated at $0$, so plug in $x = 0$ to get $27$.
For the $x$ intercepts, factor the expression $-x^3 + 27$. Note that the factorization you provided is incorrect, you forgot to distribute the negative. The correct answer would be $-(x^3 - 27) = -(x - 3)(x^2 + 3x + 9)$. Then use zero product property and quadratic formula.
