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secondary 4 | A Maths
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This is a linear law A-math question. How to make this eqn into Y = mX + C
Date Posted: 2 years ago
Views: 209
Bring the x3 over. This is a trick when there are four terms involved because in Y = mX + c there are only three terms.
And you have
Y = y - x3
X = x
m = a
c = b
And you have
Y = y - x3
X = x
m = a
c = b
See 1 Answer
y = x³ + ax + b
For the equation Y = mX + C , you need constants C and m for the y-intercept and gradient respectively. The other terms are variables.
In this equation, you only have constants a and b.
What you can do is : divide both sides by x.
y/x = x² + a + b/x
Then, y/x - x² = b/x + a
y/x - x² = b(1/x) + a
So comparing this to the above,
Y = y/x - x², X = 1/x , m = b, C = a
(Disclaimer: the line is undefined for x = 0 since division by 0 is undefined. So it will work for lines that do not cut the y-intercept
Alternatively,
Simply subtract x³ on both sides (or you can call it 'bringing' over)
y - x³ = ax + b
Here, Y = y - x³, X = x, m = a, C = b
(This line equation is defined for when x = 0
and other values of x)
For the equation Y = mX + C , you need constants C and m for the y-intercept and gradient respectively. The other terms are variables.
In this equation, you only have constants a and b.
What you can do is : divide both sides by x.
y/x = x² + a + b/x
Then, y/x - x² = b/x + a
y/x - x² = b(1/x) + a
So comparing this to the above,
Y = y/x - x², X = 1/x , m = b, C = a
(Disclaimer: the line is undefined for x = 0 since division by 0 is undefined. So it will work for lines that do not cut the y-intercept
Alternatively,
Simply subtract x³ on both sides (or you can call it 'bringing' over)
y - x³ = ax + b
Here, Y = y - x³, X = x, m = a, C = b
(This line equation is defined for when x = 0
and other values of x)
Typo : it should be 'do not cut the y-axis' instead of the 'do not cut the y-intercept'
If your data points do not include x = 0 , and only include either positive or negative values of x (x > 0 or x < 0, but not both) then the first way is okay.
Otherwise, do the second way.
If your data points do not include x = 0 , and only include either positive or negative values of x (x > 0 or x < 0, but not both) then the first way is okay.
Otherwise, do the second way.
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