Question 564909
finding the vertex of y=x^2+3x-4
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The vertex is on the LOS, the Line of Symmetry.
The LOS is x = -b/2a = -3/2
x = -3/2
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The vertex is (-3/2,f(-3/2))
f(-3/2) = (-3/2)^2 + 3*(-3/2) - 4 = 9/4 - 18/4 - 16/4
= -25/4
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Vertex at (-3/2,-25/4)
or (-1.5,-6.25)
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You can find it by putting the eqn in standard form, also.
y=x^2+3x-4 = (x^2 + 3x + 2.25) - 4 - 2.25
y = (x + 1.5)^2 - 6.25
--> vertex at (-1.5,-6.25)