Draw a graph and clearly label any x and y intercepts for the equation y=x^2+6x+9

Step 1: Factorise the equation either directly or by using the quadratic formulae {[-b(+/-)sqrt(b^2-4ac)]/2a}. This equation can be factorised directly since 3*3=9 (for the last term on the RHS) and 3+3=6 (for the second term on the RHS). Thus y=x^2+6x+9 can be written as y=(x+3)^2 | Step 2: Find intercept(s) on the x-axis – they exist when y=0, thus we have:(x+3)(x+3)=0. Since the RHS of the equation is 0, this means that (x+3) has to also equal 0 for the equation to be true, giving us x=-3. So our x-axis intercept is at the point x=-3 | Step 3: Find the y-axis intercept – they exists when x=0. Thus substitute x=0 in y=x^2+6x+9 , which gives us y=9 | Step 4: Draw the graph. Since x^2 term is positive the graph will take a u–shape ( if the x^2 term was negative the graph will take an n–shape), with the bottom end of the u-shape just touching the x-axis at x=-3 and the RHS end of the u-shape graph intercepting the y-axis at y=9.