Graph the function using it vertex, axis of symmetry and intercepts f(x)=x^2+6x+9

Question

Not sure if i have the right points.  I have a vertex of (-3,0),  axis of sym X=-3, y=9, x=3, -3.  but it doesn't look right on the graph

Answer 1

The function is a perfect square (x+3)^2, which means there's only one zero at x=-3. This is the vertex of the parabola with a line of symmetry x=-3. You can see it must be a line of symmetry because at an equal distance from either side of this line, the square of the x value is the same. For example, x=-4 and x=-2, x+3=-1 or 1 which have 1 as their square. You're right about the vertex at (-3,0); axis of symmetry is x=-3, correct again; the only root is at the vertex; the y intercept is 9, as you worked out. It was the x intercepts (the curve touches but doesn't intercept the x axis) that let you down, because x=3 is not a zero (y=36).