vertex (y) of (-1/2, f[-1/2])?
I'm studying for a midterm and I'm stumped on this one. The original question is....f(x)= -x^2(squared) -x+6.....i understand that the x(axis of symmetry) is -1/2....but why is the vertex (-1/2, 25/4)? I keep getting (-1/2, 27/4)
Here's what I did:
I used the equation -b/2a to get the axis of symmetry...so that's 1/2(-1) and got -1/2...so to get the vertex I used f(-1/2)= 1/2^2(squared)+1/2+6 and get 1/4+1/2+6= 3/4+6= 3/4+24/4= 27/4! someone please help! thanks!
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Verified answer
-x^2 in the original equation is - (x^2), not (-x)^2
therefore, f(-1/2)
= - (-1/2)^2 - (-1/2) + 6
= - (1/4) + 1/2 + 6
= 1/4 + 6
= 25/4
you have to square x, then multiply by -1
what you did was to multiply by -1, then squared the answer giving a value of,
1/4 + 1/2 + 6
= 27/4
this is wrong
answer for f(-1/2) is 25/4