Solve `y=x^2+4x+4` and `8x+4y+15=0` simultaneously using graphing.
(1) The graph of `y=x^2+4x+4=(x+2)^2` is a parabola, opening up, with vertex at (-2,0).
(2) 8x+4y+15=0 is a line; in slope-intercept form it is y=-2x-15/4.
Plot both graphs and estimate where they intersect.
The intersections appear to be somewhere left of x=-4 and to the right of x=-2.
A good approximation is `x~~-4.1` and `x~~-1.9` . Plug these values into the equations to get the y-values if needed.
Thus the intersections are approximately (-4.1,4.45) and (-1.8,-.15).
** If you are interested, the actual x-values of the intersections are `x=-3+-(sqrt(5))/2` . **