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Why does the translation y=f(x+2) translate the graph f(x) 2 units left instead of 2 to the right?

It seems like f(x+2) should translate the graph y=f(x) 2 to the right, it is +2 inside the bracket after all? However, if we think of the translation as actually subbing in x+2 into the equation y=f(x), it m...
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Answered by Seth H. • Maths tutor
12049 Views

Express 4x/(x^2-9)-2/(x+3) as a single fraction in its simplest form

4x/(x 2 -9)-2/(x+3) = 4x/((x+3)(x-3))-2/(x+3) = 4x/((x+3)(x-3))-2(x-3)/((x+3)(x-3)) = (4x-2(x-3))/((x+3)(x-3)) = (2x+6)/((x+3)(x-3) = (2(x+3))/((x+3)(x-3)) = 2/(x-3)
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Answered by Charles S. • Maths tutor
9179 Views

Find the equation of the normal to the curve y=2x^3 at the point on the curve where x=2. Write in the form of ax+by=c.

For x = 2, y = 16. Calculate the gradient of the curve at y = 2, dy/dx = 6x^2, dy/dx = 24. This is also the gradient of the tangent to the curve at x = 2. It is a rule that the products of the gradients of t...
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Answered by Maddy L. • Maths tutor
10239 Views

Given that y=π/6 at x=0 solve the differential equation,dy/dx=(e^x)cosec2ycosecy

This question challenges a students skills with integration and trigonometry a very common question on Core 4 papers that can leave students struggling. First we use our knowledge that cosecy=1/siny and mult...
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Answered by Oliver L. • Maths tutor
13747 Views

Find the area enclosed between C, the curve y=6x-x^2, L, the line y=16-2x and the y axis.

First we need to find the intersection point(s) of L and C so set 6x-x 2 =16-2x and rearrange to get x 2 -8x+16=0 so (x-4) 2 =0.Repeated root so line is tangent to the curve at x=4, y=16-2(4)=8 that is the p...
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Answered by David M. • Maths tutor
4180 Views