Question 867079
Given equation {{{highlight_green(x=(y+2)^2)}}}


<i>1. Show the five points of the graph and how you got them.</i>


What five points?  The graph of the equation has infinite number of points.  


<i>2. what are the start/end points?</i>


What do you mean?  This graph has a vertex on the left and infinite number of points to the right.


<i>3. what is the general shape and location of the graph?</i>


The equation is in standard form for a parabola with a horizontal axis of symmetry.  The parabola opens toward the right and the vertex (based on knowing how to read from the standard form equation) is (0, -2).  The graph is symmetric around {{{y=-2}}}.

<i>4. State the domain and range for equation in interval notation.</i>


The point farthest to the left is (0, -2).  The graph and equation are of TWO <b>separate</b> functions.  The domain for each branch is {{{0<=x}}} ( using inequality relationship notation, and not as you asked in interval notation ).  The range for the upper branch is {{{y>=-2}}} and the range for the lower branch is {{{y<=-2}}}.


<i>5. State whether the equation is a function or not giving your reason why.</i>


The equation is NOT a function.  Try the "Vertical Line Test".  What does it tell you?  Look at any value of x in the domain of {{{y=(x+2)^2}}}.  If you find more than one value for y for ANY value of x, then the relation is not a function.  A function must have no more than one value for output for any input.