Andy Dufresne
Badges: 4
Rep:
?
#1
Report Thread starter 1 month ago
#1
Hi! I was wondering if anyone could help out with this maths problem from the Khan Academy – I got it wrong and don’t understand the answer explanation it gives:

‘The path of an atom where x is the east coordinate (in mm) and y is the north coordinate (in mm) from a sensor is:
X=y² -4y +5
Among all points on the atom's path, what is the smallest east coordinate in mm?’

The Answer Explanation:
‘We can find the smallest east coordinate by rewriting the quadratic equation in vertex form:
x = a(y - y₀)² + x₀
where ‘a’ is a real number and (x₀, y₀) is the vertex of the parabola formed and, therefore, the minimum or maximum.
We rewrite the expression in vertex form by completing the square:
x = y² -4y +5
x = (y – 2)² + 1
Because x is the dependent variable and a is positive (it is equal to 1), x achieves a maximum value of x₀ = 1. The smallest east coordinate for the path of the atom is 1 millimetre.’

I thought the answer would be 2, because in my own notes I have the vertex form as being
Y = a(x - h)² +k with the vertex’s coordinate being (h,k). But in this answer explanation, even though they’re using slightly different symbols, (x₀, y₀), they seem to have put the coordinates the other way around?
Is there a particular reason for this? I suspect I must be missing something massive here, rather than the KA having botched up the question.
Thanks in advance! 
0
reply
mqb2766
Badges: 18
Rep:
?
#2
Report 1 month ago
#2
Sure, the quadratic is modified (reflection in y=x) so you now have
x = y^2 ...
rather than
y = x^2 ...
So completing the square in y (the quadratic term) gives the minimum value for x and the y value which generates it.
Last edited by mqb2766; 1 month ago
0
reply
Andy Dufresne
Badges: 4
Rep:
?
#3
Report Thread starter 1 month ago
#3
(Original post by mqb2766)
Sure, the quadratic is modified (reflection in y=x) so you now have
x = y^2 ...
rather than
y = x^2 ...
So completing the square in y (the quadratic term) gives the minimum value for x and the value of which which generates it.
thanks. so when its a modified quadratic in vertex form the vertex is actually (k,h) ?
0
reply
mqb2766
Badges: 18
Rep:
?
#4
Report 1 month ago
#4
(Original post by Andy Dufresne)
thanks. so when its a modified quadratic in vertex form the vertex is actually (k,h) ?
Just think of completing the square in whichever variable has the quadratic term so
y = x² - 4x +5
gives
y = (x-2)^2 + 1
as normal and the quadratic is a bowl with minimum (in y) at (x,y) = (2,1)

x = y² - 4y +5
gives
x = (y-2)^2 + 1
as normal and the quadratic is a "c" shape with minimum (in x) at (x,y) = (1,2).


Its just a reflection in the line y=x (or simply swap the variables / values).
0
reply
Andy Dufresne
Badges: 4
Rep:
?
#5
Report Thread starter 1 month ago
#5
(Original post by mqb2766)
Just think of completing the square in whichever variable has the quadratic term so
y = x² - 4x +5
gives
y = (x-2)^2 + 1
as normal and the quadratic is a bowl with minimum (in y) at (x,y) = (2,1)

x = y² - 4y +5
gives
x = (y-2)^2 + 1
as normal and the quadratic is a "c" shape with minimum (in x) at (x,y) = (1,2).


Its just a reflection in the line y=x (or simply swap the variables / values).
I think I get it now - thanks so much for the help ! :-)
1
reply
mqb2766
Badges: 18
Rep:
?
#6
Report 1 month ago
#6
(Original post by Andy Dufresne)
I think I get it now - thanks so much for the help ! :-)
https://www.desmos.com/calculator/344jpmraeb
1
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

How do you feel about your grades? Are they...

What I expected (55)
24.77%
Better than expected (49)
22.07%
Worse than expected (118)
53.15%

Watched Threads

View All