# maths question

#1
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
0
4 months ago
#2
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?
1
#3
(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
0
4 months ago
#4
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.
0
#5
(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
0
4 months ago
#6
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.
0
#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
0
4 months ago
#8
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?
Last edited by mqb2766; 4 months ago
0
#9
(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
Last edited by daisychad98; 4 months ago
0
4 months ago
#10
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.
Last edited by mqb2766; 4 months ago
0
#11
(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
1
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