\[f(x)=4(x-2)^{2}-1\] vertex?

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\[f(x)=4(x-2)^{2}-1\] vertex?

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Have you tried graphing? :)
(2, -1)
(2,-1)

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Other answers:

nope not yet
it is displayed for you.
you don't have to graph this to find the vertex it is (2,-1)
\[f(x)=a(x-h)^2+k\] has vertex (h,k)
i was just fixing to say that :(
what would the x-intercept of parabola for this would be??
set y=0 and solve for x
????

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