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The circle C has centre (3, 1) and passes through the point P(8, 3). (a) Find an equation for C. (b) Find an equation for the tangent to C at P, giving your answer in the form ax + by + c = 0 , where a, b and c are integers.

A)1) Draw a diagram of the circle displaying the centre and perimeter points along with their respective co-ordinates.2) Write down the equation for a circle labelling the centre and perimeter points. 3) Inp...
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Answered by Patrick V. • Maths tutor
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How would you integrate ln(x)

It can seem tricky to integrate ln(x), as there is no obvious solution to do it.It is, however, quite simple to do if you use the 'by parts' method.If you have y=ln(x)Set u=ln(x) and dv/dx=1That gives du/dx=...
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Answered by Adam C. • Maths tutor
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y=e^2x-11e^x+24 Find the stationary point, nature of the stationary point, the x-intercepts and the y-intercept (calculator allowed)

Stationary point: dy/dx = 2e^2x - 11e^x =0 2e^2x = 11e^x e^x=5.5 (can divide by e^x since e^x > 0 for all x) x=ln(5.5), y=5.5^2-11 5.5+24=-6.25 Answer: (ln(5.5),-6.25) Nature of stationary point : Evaluat...
AJ
Answered by Asmita J. • Maths tutor
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What is the gradient of this curve y=5x^3+6x^2+7x+8 at point x=3?

When differentiating an equation (y) you find the equation of the gradient, called dy/dx. The rule for differentiating a power of x is given below:y=x^n dy/dx= nx^(n-1)Applying this rule to this question you...
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Answered by Tutor179115 D. • Maths tutor
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A curve has an equation y=3x-2x^2-x^3. Find the x-coordinate(s) of the stationary point(s) of the curve.

The very first step in solving this problem is understanding that a stationary point is where the derivative of the curve, dy/dx (or in Newton’s notation y’), is equal to zero. This is because at stationary ...
CG
Answered by Callum G. • Maths tutor
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