Question 1159758
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Put the equation in vertex form: {{{y = a(x-h)^2+k}}}<br>
In that form, the vertex is at (h,k), and the axis of symmetry is x=h.<br>
Complete the square....<br>
{{{x^2+6x+3 = (x^2+6x+9)+3-9 = (x+3)^2-6}}}<br>
This form of the equation tells us that the vertex is at (-3,-6) and the axis of symmetry is x=-3.  That can be seen in the graph:<br>
{{{graph(400,400,-10,10,-10,10,x^2+6x+3)}}}<br>
For the y-intercept, set x=0 and evaluate.  Using the given form of the equation, it is easy to see that the y-intercept is (0,3).<br>
The y-intercept is 3 units to the right of the axis of symmetry; the point symmetric with the y-intercept will be 3 units to the left of the axis of symmetry, and it will have the same y value as the y-intercept: (-6,3).  That is also easily seen in the graph.<br>