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secondary 4 | A Maths
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Date Posted: 4 years ago
Views: 375
1/y = mx² + c
Sub x = 1, y = ½,
1/(½) = m(1²) + c
2 = m + c
m = 2 - c
Sub x = 2, y = ⅓,
1/(⅓) = m(2²) + c
3 = 4m + c
Since m = 2 - c,
3 = 4(2 - c) + c
3 = 8 - 3c
3c = 5
c = 5/3
m = 2 - 5/3
= ⅓
Sub x = 3, m = ⅓, c = 5/3,
1/y = ⅓(3²) + 5/3
1/y = 14/3
y = 3/14
Sub x = 1, y = ½,
1/(½) = m(1²) + c
2 = m + c
m = 2 - c
Sub x = 2, y = ⅓,
1/(⅓) = m(2²) + c
3 = 4m + c
Since m = 2 - c,
3 = 4(2 - c) + c
3 = 8 - 3c
3c = 5
c = 5/3
m = 2 - 5/3
= ⅓
Sub x = 3, m = ⅓, c = 5/3,
1/y = ⅓(3²) + 5/3
1/y = 14/3
y = 3/14
Thanks J :D
Welcome
Yup this approach is good cause gradient is always constant. And we don’t have to trouble ourselves to find the equation of line!
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done
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We know that equation of straight line is y=Mx+c. Since x^2 and 1/y plot to form a straight line. We can substitute those as our x and y. Hope this helps!
Date Posted:
4 years ago
Thank you! Wait but isnt m+c=2, not 1?
Oh yes sorry for the mistake. Just make the adjustments from there should be able to arrive at the correct equation. Sorry about that!
Small error there.
Thank you for the help! I was stumped.
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