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Scottish Highers

Find ∫((x^2−2)(x^2+2)/x^2) dx, x≠0

Before we start any integration steps it would be wise to try and simplify the fraction that we have been given. We want to get rid of that denominator so we will start by expanding out the numerator. Thi...

Answered by Anthony S. • Maths tutor

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A circle has equation x^2+y^2+6x+10y-7=0. Find the equation of the tangent line through the point on the circle (-8,-1).

The circle can be written (x+3)²+(y+5)^2 = 41 and so has center (-3,-5).
The gradient through the center and the point (-8,-1) is;m = (-5+1)/(-3+8) = -4/5.
The tangent line is perpend...

Answered by Rhianna L. • Maths tutor

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y=x^3-3x^2+2x+5 a)Write down the coordinates of P the point where the curve crosses the x-axis. b)Determine the equation of the tangent to the curve at P. c)Find the coordinates of Q, the point where this tangent meets the curve again.

a) P (0,5)b) y = 2x + 5c) Q (3,11)

Answered by Harrison A. • Maths tutor

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dy/dx = 6x^2 - 3x + 4 when y=14 x=2 Find y in terms of x

∫dy/dx = ∫6x^2 - 3x + 4y = (6/3)x^3 - ( 3/2)x^2 + 4x + cLet y = 14 and x = 214 = 2(2^3) - (3/2)(x^2) + 4(2) + c14 = 2(8) - (3/2)(4) + 4(2) + c14 = 16 - 6 + 8 + c14 = 18 + cc = -4therefore:y = 2x^3 - (3/2...

Answered by Lauren A. • Maths tutor

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PQR is a triangle with vertices P (−2, 4), Q(4, 0) and R (3, 6). Find the equation of the median through R.

(1) Find the Midpoint of PQ which is (1,2) (Halfway between the x and y coordinates)(2) dy/dx for M(1,2) -> R(3,6) = (6-2)/(3-1) = 4/2 = 2(3) y =mx + c so y = 2x + c when R(3,6) is input 6 = 2(3) + c, ...

Answered by Scott H. • Maths tutor

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