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Maths
A Level

A curve, C, has equation y =(2x-3)^5. A point, P, lies on C at (w,-32). Find the value of w and the equation of the tangent of C at point, P in the form y =mx+c.

To find the value of w, let x = w and y = -32. Substitute these values into the equation of the curve, C: y = (2x-3)^5 => -32 = (2(w) - 3 )^5. Note: the symbol, =>, means "implies that." F...

LM
Answered by Lewis M. Maths tutor
3627 Views

A curve has equation y = f(x) and passes through the point (4, 22). Given that f ′(x) = 3x^2 – 3x^(1/2) – 7, use integration to find f(x), giving each term in its simplest form.

Firstly we can use the difference rule to split f'(x) into three components which we can consider separately. Then using the knowledge that the integral of x^n is 1/(n+1)*x^(n+1) we get the expression for...

AS
Answered by Abbey S. Maths tutor
3512 Views

y = 4x/(x^2+5). a) Find dy/dx, writing your answer as a single fraction in its simplest form. b) Hence find the set of values of x for which dy/dx < 0

a) We need to differentiate this equation using the quotient rule (Given that it is a fraction with an x term on both the top and bottom of the fraction). We assign the numerator and denominator as follow...

JF
Answered by James F. Maths tutor
6885 Views

A curve has parametric equations: x=(t-1)^3 and y= 3t - 8/(t^2). Find dy/dx in terms of t. Then find the equation of the normal at the point on the curve where t=2.

dx/dt = 3(t-1)2dy/dt = 3 + 16t-3dy/dx=(dy/dt)(dt/dx) dy/dx = 3 + 16t-3 / 3(t-1)2
At t=2 dy/dx= (3 + 16/8) / 3 = 5/3 Gradient of the normal = -3/5with t=2 y...

JH
Answered by Jasmin H. Maths tutor
3334 Views

How can functions be transformed?

A function, y = f(x), with y on the vertical axis and x on the horizontal axis, can be transformed by 3 different ways: It can be stretched (or shrunk)If y = f(ax), the function is stretc...

JM
Answered by Jack M. Maths tutor
2810 Views

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