How do you write the quadratic in vertex form given #y=2x^2+3x-8#?

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Answer 1 · 2015-04-29T07:33:20

The vertex form of a quadratic function is given by
#y = a(x - h)^2 + k#, where #(h, k)# is the vertex of the parabola.

We can use the process of Completing the Square to get this into the Vertex Form.

#y=2x^2+3x-8#

#-> y + 8 = 2x^2 +3x# (Transposed -8 to the Left Hand Side)

#-> y + 8 = 2(x^2 + (3/2)x)# (Made the coefficient of #x^2# as 1#

Now we add #2*(3/4)^2# to each side to complete the square

#-> y + 8 + 2*(3/4)^2 = 2(x^2 + (3/2)x + (3/ 4)^2)#

#-> y + 8 + 9/8 = 2(x+3/4)^2 #

# -> y + 73/8 = 2{x - (-3/4)}^2#

# -> color(green)(y = 2{x - (-3/4)}^2 + (-73/8)# is the Vertex Form

The vertex of the Parabola is# {-3/4 , -73/8}#