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maths question

hello I was wondering if anyone could help me with this question
find the coordinates of the turning point on the curve with equation y=9+18x-3x^(2)
(4 marks)
thank you
gcse level higher non caculator
Reply 1
Avatar for mqb2766
mqb2766
21
Original post by daisychad98
hello I was wondering if anyone could help me with this question
find the coordinates of the turning point on the curve with equation y=9+18x-3x^(2)
(4 marks)
thank you
gcse level higher non caculator

have you tried completing the square?
Reply 2
Avatar for daisychad98
daisychad98
OP
7
Original post by mqb2766
have you tried completing the square?

yes, i did
y = -3( x^2 - 6x - 3)
y = -3(( x-6)^2- 36-3)
y = -3(( x-6)^2- 39)
y = -3( x-6)^2- 117

than wasn't sure what to do next
Reply 3
Avatar for mqb2766
mqb2766
21
Original post by daisychad98
yes, i did
y = -3( x^2 - 6x - 3)
y = -3(( x-6)^2- 36-3)
y = -3(( x-6)^2- 39)
y = -3( x-6)^2- 117

than wasn't sure what to do next

The bold part isn't correct. Try expanding the x term in your answer you get 36x.

But when you complete the square, what do the terms represent in terms of a turning point.
Reply 4
Avatar for daisychad98
daisychad98
OP
7
Original post by mqb2766
The bold part isn't correct. Try expanding the x term in your answer you get 36x.

But when you complete the square, what do the terms represent in terms of a turning point.

y = -3( x^2 - 6x - 3)
y = -3(( x-3)^2- 9-3)
y = -3(( x-3)^2- 12)
y = -3( x-6)^2+36
ohh
so the coordinates are
(6,36)
i think the number inside the bracket is x and the number outside the bracket is y
and because x is inside the bracket you need to change the sign
thank you
Reply 5
Avatar for mqb2766
mqb2766
21
Original post by daisychad98
y = -3( x^2 - 6x - 3)
y = -3(( x-3)^2- 9-3)
y = -3(( x-3)^2- 12)
y = -3( x-6)^2+36
ohh
so the coordinates are
(6,36)
i think the number inside the bracket is x and the number outside the bracket is y
and because x is inside the bracket you need to change the sign
thank you

Nearly there, see the bolds.
If you're unsure, just expand the completed square and check it matches the original quadratic.
Reply 6
Avatar for daisychad98
daisychad98
OP
7
Original post by mqb2766
Nearly there, see the bolds.
If you're unsure, just expand the completed square and check it matches the original quadratic.

oh yes, silly mistake
thank you
y = -3( x-3)^2+36
so it would be 3,36 ?
or do i have to times -3 by -3 first
Reply 7
Avatar for mqb2766
mqb2766
21
Original post by daisychad98
oh yes, silly mistake
thank you
y = -3( x-3)^2+36
so it would be 3,36 ?
or do i have to times -3 by -3 first

Agree for the (x,y) point. A simple way to understand it is to sub x=3 into the expression.
What happens to the quadratic term? Why does this mean
y <= 36
in this case. What does the -3 multiplier affect?
(edited 2 years ago)
Reply 8
Avatar for daisychad98
daisychad98
OP
7
Original post by mqb2766
Agree for the (x,y) point. A simple way to understand it is to sub x=3 into the expression.
What happens to the quadratic term? Why does this mean
y <= 36
in this case.
9+18x-3x^(2) = y

9+18x-3x^(2) = y
9+(18x3)-3^2x3=y
y=36
I'm not sure what happens to the quadratic term ?
thank you
(edited 2 years ago)
Reply 9
Avatar for mqb2766
mqb2766
21
y(3) = -3(3-3)^2 + 36 = -3*0 + 36 = 36
x=3 makes the quadratic term (x-3)^2 zero, so the quadratic is at a turning/stationary/extreme point. The value of the quadratic is simply the last term in the completed square expression. For x != 3, then
-3(x-3)^2 < 0
as the quadratic part is > 0 and multiplying it by -3 makes it < 0. So
y <= 36
The -3 multiplier affects the curvature of the quadratic, not the value of the turning point.
(edited 2 years ago)
Original post by mqb2766
y(3) = -3(3-3)^2 + 36 = -3*0 + 36 = 36
x=3 makes the quadratic term (x-3)^2 zero, so the quadratic is at a turning/stationary/extreme point. The value of the quadratic is simply the last term in the completed square expression. For x != 3, then
-3(x-3)^2 < 0
as the quadratic part is > 0 and multiplying it by -3 makes it < 0. So
y <= 36
The -3 multiplier affects the curvature of the quadratic, not the value of the turning point.

oh that makes sense so it doesn't change the turning point
thank you for your help

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