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Q14 (1 Viewer)

crazy_paki123

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the answer is k<-32, you draw a horizontal lone (k=-32) which is the last time the line intersects the quartic as it is the min y value thus under it there are no graphical solutions due to no intersections. i did a similar question with my tutor the day before the exam :L
 

gr_111

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the answer is k<-32, you draw a horizontal lone (k=-32) which is the last time the line intersects the quartic as it is the min y value thus under it there are no graphical solutions due to no intersections. i did a similar question with my tutor the day before the exam :L
sack your tutor mate, its definitely k>32.
 

narenkumar

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Did we have to write it as a fraction.
I wrote it as a decimal. How much mark will I lose.
 

lance687876

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the graph didn't work for me, l got a min at 0,0 and a max at -1,-5 and a max at -2,-32 so l just... drew it l guess o_O
l'm guessing l got it wrong so that's 20 marks gone in total.
wow.
never gonna underestimate an exam again.
stupid silly mistakes.
 

Timske

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the graph didn't work for me, l got a min at 0,0 and a max at -1,-5 and a max at -2,-32 so l just... drew it l guess o_O
l'm guessing l got it wrong so that's 20 marks gone in total.
wow.
never gonna underestimate an exam again.
stupid silly mistakes.
(1,-5) min
(0 , 0) max
(-2,-32) mini

thats what i got
 

RishBonjour

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it looked like a normal quartic with an upside down parabola in the middle
 

Timske

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how did u get 0,0 as a maximum i couldnt differentiate it as a maximum or minimum so i wrote point of inflection T_T
Well there are two ways you could do this; differentiate again or simply sub in values to the left and right of 0.
 

baileychen

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Well there are two ways you could do this; differentiate again or simply sub in values to the left and right of 0.
what was the equation again i remember differentiating again and getting 0 might have been a silly mistake
 

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