Question 767627
Here is the graph of y=(x+2)^2-1
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{{{ graph( 300, 200, -5, 5, -5, 5, (x+2)^2-1) }}}
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vertex is (-2, -1)
axis of symmetry is x = -2
parabola opens upward
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find focus point
y=(x+2)^2-1,***for upward pointing parabola x^2 = 4ay
(4/4)y = (x+2)^2-1
factor the 4 out and we have
4*(1/4)y = (x+2)^2-1
focus is at (-2, -3/4)
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latus rectum is the line through the focus perpendicular to the axis of symmetry, in this case it is the line y = -3/4