ATAR Notes: Forum
HSC Stuff => HSC Maths Stuff => HSC Subjects + Help => HSC Mathematics Advanced => Topic started by: Albertenouttaten on March 14, 2021, 10:08:49 pm
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Hi all,
Been absolutely stumped on the following question. Help really is just needed on the first sub-question. I've tried using vertex form and standard form of parabola through substituting, however still not able to do it. Any help would be appreciated.
Regards
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If the point passes through (0, 8 ), (7.5, 0) and (-7.5, 0):
- We clearly have a cap parabola ie. the coefficient on the \(x^2\) term is negative
- \(f(0) = 8 \implies f(x) = ax^2 + bx + 8\) for some constants \(a, b\) (reminder that we also know \(a < 0\))
- \(f(7.5) = f(-7.5) \implies b = 0\) from above (if a function is even, then all the coefficients of odd powered terms are zero - you could also surmise that \(f(x) - f(-x) = 0 \implies 2bx = 0\))
- Hence we basically have to solve \(0 = a(7.5)^2+8 \implies a = -\frac{32}{225}\)
Hence our equation is just \(f(x) = -\frac{32}{225}x^2 + 8\)
Hope this makes sense :)
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Hi, understand all but the second line ("from above (if a function is even, then all the....").
Also, really appreciate the help so far!
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That's a bit of a nice result that I find somewhat interesting (and you might as well! - it may or may not be something that has been covered). The more important bit is that \(f(7.5) = f(-7.5) \implies f(7.5) - f(-7.5) = 0\) ie. \(2b(7.5) = 0\) which implies that b = 0. It's a collection of suggestions you could follow to find out that b = 0 - it's not important that you understand everything in that line, just that you know how we can conclude that b = 0.