Question 18723
The instructions are to solve the inequality. Find the vertex and intercepts, then graph it:
y = -x^2 - 3x - 2
y=-(x^2+3x+2)=-(x^2+2x+x+2)=-(x(x+2)+1(x+2))=-(x+1)(x+2)
we first analyse this ..that is what will be its value if x=0,when will y be zero ,when is it positive,when is it -ve,has it got a maximum or minimum etc...
we can give different values for x and find the corresponding values of y to graph it..let us do 
let us put x=0....1....2....3....-1....-2....-3....to show you the method
we get.....y=-2...-6...-12..-20...0.....0....-2....etc..now you can plot a graph with this table...you can take more points also and do it.. 
{{{ graph( 150, 100, -6, 6,-60,5,-x^2-3x-2) }}} 
we refer to y as the function of x given by y=f(x)= -x^2 - 3x - 2."it"refers to function or y or f(x)
we find it has two zeros (that is y=0)at x=-1 and x=-2...
it is  +ve when x is between -1 and -2.
it is -ve if x<-2...or if x>-1
its x intercepts are (that is value of x when y is zero )-1 and -2..that is the graph cuts x axis at x=-1 and x=-2
its y intercept is ( that is value of y when x is zero) is -2
its vertex is at the point when y is minimum or maximum..for this we write y as follows
y=-{{{(x+(3/2))^2-(2-9/4)}}}=-{{{(x+(3/2))^2+(1/4)}}}
now we reason that {{{(x+(3/2))^2}}}being perfect square ..it is always positive.hence its minimum value is zero (when x=-3/2)..then y =1/4
so we call x=0 and y=1/4 as the vertex of the graph.