K-1972
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I have only just realised that i don't know how to draw a quadratic with no real roots on a graph. I had the co ordinate of the point where the curve crossed the y axis (0,11) but when i came to draw the curve, the mark scheme said i was wrong because i drew the minimum point of the curve in the wrong quadrant. How do i work out which quadrant the minimum/maximum point is on?
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The quadrant is just implied by the minimum point itself, for example if the minimum point is (-2, 1) you know it'll be in the top left quadrant.
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K-1972
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(a) Show that x^2 + 6x + 11 can be written as (x+p)^2 +q


where p and q are integers to be found.
(b) Sketch the curve with equation y = x^2 + 6x + 11, showing clearly any intersections with

the coordinate axes.
(c) Find the value of the discriminant of x^2 + 6x + 11.



here is the question. i can do a and c but it was b where i messed up. it doesn't give you the minimum point
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RDKGames
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(Original post by oOKathrynOo)
(a) Show that x2 + 6x + 11 can be written as (x+p)2 +q


where p and q are integers to be found.
(b) Sketch the curve with equation y = x2 + 6x + 11, showing clearly any intersections with

the coordinate axes.
(c) Find the value of the discriminant of x2 + 6x + 11.


here is the question. i can do a and c but it was b where i messed up. it doesn't give you the minimum point
The completed square form gives the minimum point which is here (-p,q)
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K-1972
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(Original post by RDKGames)
The completed square form gives the minimum point which is here (-p,q)
ohhhhh ok thanks a lot. feel much more confident going into the exam tomorrow
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sulaimanali
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can u give the full questions plz?
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Shadowfire123
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(Original post by sulaimanali)
can u give the full questions plz?
It is the gold 3 paper Q3 ojn physicsandmathstutor


(Original post by oOKathrynOo)
I have only just realised that i don't know how to draw a quadratic with no real roots on a graph. I had the co ordinate of the point where the curve crossed the y axis (0,11) but when i came to draw the curve, the mark scheme said i was wrong because i drew the minimum point of the curve in the wrong quadrant. How do i work out which quadrant the minimum/maximum point is on?
Check your private message.. I have replied..hope you understood it.
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sulaimanali
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thanks
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(Original post by sulaimanali)
thanks
Yes! Good luck
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(Original post by Shadowfire123)
Yes! Good luck
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(Original post by Redcoats)
1) +C
2)a=1/16
3) a2 = 2k
a3 = k+1/2
k=17/6
4) d=6
10316 total
5) x=3 and x=-1
6) (x-4)^2+3
Minimum point at (4,3) intercept at (0,19)
PQ : 4sqrt(17) units
7)
8) 5x-4y-49=0
Area = 369/10
9) line theough (0,C)
Asymptote at y= 5
equate and disciminant greater than 0
c<5-2sqrt(3) and c> 5+sqrt(3)
10) k = 75
c = 5/2
just differentiate
x = -5/3

(some questions might be missing; reply with edits)
It was a reasonable paper- easier than last year - how did you guys find it?
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