How to graph a parabola #y=x^(2)-2x-15#?

1 Answer
Skewd
Jun 21, 2018

You have two choices
1. complete the square to get standard form, find vertex and 2 other points.
2. Use the equation #x=(-b/2a)# to find the x-coordinate of the vertex.

Explanation:

Complete the square...
#y=(x^2-2x+(b/2)^2)-15-(b/2)^2#
#y=(x^2-2x+1)-15-1#
#y=(x-1)^2-16# so the vertex is at (1, -16)
find two more points, use x-values near the vertex.
(0, -15) and by symmetry (2, -15) and (-1, -12).

Alternatively,
Use #h=(-b/2a)# for the x-coordinate of the vertex
and #k=# the y-value when you substitute x = h into the equation.
write as #y=a(x-h)^2+k# here a=1, the vertex is (h, k).
You need to calculate 2 more points as above.

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