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Fill in the missing term so that the quadratic equation has a graph that opens up, with a vertex of (– 2, – 16), and x intercepts at x = -6 and x = 2. (Do not include the negative sign in your answer.) y = x2 + 4x − ___
whats the question nvm
vertex form \[y=a(x-h)^2+k\]where the vertex is (h,k) so we have \[y=a(x+2)^2-16\]we know that when x = -6 y = 0 \[0=a16-16 \\so\\a=1\]we check now that it works for our other point (2,0) \[0=a(2+2)^2-16\\so\\a=1\\so\\y=(x+2)^2-16\]
\[y=x^2+4x+4-16=x^2+4x-8\]
soo im kinda confused. give me a min.
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nvm
do you understand it?
yes thank you
np
@zzr0ck3r can you helpwith two more?
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