Question 982200
What do you want?  Solve, graph, general into standard form, intercepts?


response:  most of it.



MISTAKE IS NOW FIXED  (it was a mistake in sign.)


COMPLETE REDO----....

{{{y=3-15x-3x^2}}}
{{{y=-3(-1+5x+x^2)}}}
{{{y=-3(x^2+5x-1)}}}
Term to complete the square is  {{{(5/2)^2=25/4}}}.
{{{y=-3(x^2+5x+(5/2)^2-25/4-4/4)}}}
{{{y=-3((x+5/2)^2-29/4)}}}
{{{highlight(y=-3(x+5/2)^2+87/4)}}}------standard form equation, so now easier to graph.


This is a parabola, concave downward with vertex as a maximum.
The leading coefficient is NEGATIVE, telling you that the vertex will be a maximum.  (the negative 3).
Vertex is  (-5/2,87/4), which is read directly from the standard form equation.


ROOTS OR ZEROS


{{{-3(x+5/2)^2=-87/4}}}
{{{(x+5/2)^2=87/(3*4)}}}
{{{x+5/2=0+- sqrt(87/3*4)}}}
{{{x+5/2=0+- (1/2)sqrt(29)}}}
{{{highlight(x=-5/2+- (1/2)sqrt(29))}}}-------the roots, zeros, x-intercepts


y-AXIS INTERCEPT
where x=0
{{{y=3-15x-3x^2}}}-----back to the original equation because this will be easier.
{{{y=3-15*0-3*0^2}}}
{{{highlight(y=3)}}}--------the y-axis intercept


<s>Graph should look like this:</s>
NOTE: This graph is coded correctly but is not displaying correctly, so 
I am not including the rendering tags.  Use a graphing tool or such software to
see the graph properly.
graph(300,300,-10,10,-3,24,-3x^2-15x+3)
somewhat different scaling for x versus y axes