What is the equation of the parabola that has a vertex at # (4, 2) # and passes through point # (6,34) #?

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Answer 1 · 2018-04-25T11:22:34

#y = 8(x-4)^2+2#

When the parabola has a vertex at #(4,2)# its equation looks like #y= a(x-4)^2+2# and we plug in #(6,34)# to find #a#:

#34=a(6-4)^2+2 #

# 32=4a #

# a=8#

So we get

#y = 8(x-4)^2+2#

We could expand this into standard form, but at this point we've answered the question so let's stop.

Check: The vertex is correct by construction.

# 8(6-4)^2 +2 = 8(4)+2 = 34 quad sqrt#