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secondary 4 | A Maths
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Need help with 9a, b and c . Preferably with clear explanations and workings. Thankyou !
Date Posted: 4 years ago
Views: 363
By 4 am if no one else has done them yet
9(a)
on y-axis, x= 0
y = 2x² - 5x + 1 = 1
pt is (0, 1)
dy/dx = 4x - 5
at x= 0, gradient is -5
.
on y-axis, x= 0
y = 2x² - 5x + 1 = 1
pt is (0, 1)
dy/dx = 4x - 5
at x= 0, gradient is -5
.
9(b)
on x-axis, y= 0 => x = 4
y = 1 - 4/x
dy/dx = 4/x²
pt is (4, 0)
at x= 4, gradient is 4/16 = 0.25
.
on x-axis, y= 0 => x = 4
y = 1 - 4/x
dy/dx = 4/x²
pt is (4, 0)
at x= 4, gradient is 4/16 = 0.25
.
(c)
y = x
x = (x+2)/x
x² = x + 2
x = 2 or -1
pts r (-1, -1) & (2, 2)
y = 1 + 2/x
dy/dx = -2/x²
at x= -1, gradient is -2/1 = -2
at x= 2, gradient is -2/4 = -0.5
.
.
.
y = x
x = (x+2)/x
x² = x + 2
x = 2 or -1
pts r (-1, -1) & (2, 2)
y = 1 + 2/x
dy/dx = -2/x²
at x= -1, gradient is -2/1 = -2
at x= 2, gradient is -2/4 = -0.5
.
.
.
See 1 Answer
done
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Something like this would be good. We just simply need to obtain the expression for dy/dx in terms of x and then find the numerical value of dy/dx at each required point.
Date Posted:
4 years ago
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