How do I use the vertex formula to determine the vertex of the graph for #2x^2+4x-5#?

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Answer 1 · 2015-05-01T00:34:17

We need to convert #y = 2x^2+4x-5# into its vertex form:
#y = m(x-a)^2+b#
(which will then provide the vertex as #(a,b)#)

#y=2x^2+4x-5#

#y=2(x^2+2x) -5#

#y=2(x^2+2x+1) -5 -2#

#y=2(x+1)-7#

#y=2(x-(-1))+(-7)#

The vertex is at #(-1,-7)#

Answer 2 · 2015-05-01T01:48:34

There is another way.
Coordinates of vertex:
#x = -b/(2a)#
#y = f[-b/(2a)]#.

In this example:

#x = -b/(2a) = -4/4 = -1#

#f(-1) = 2(-1)^2 + 4(-1) - 5 = 2 - 4 - 5 = -7#

Answer: Vertex (-1, -7)