*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` . **

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