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How do we work out the asymptotes of the graph y=1/x -5

In most core 1 papers this kind of question is usually asked. First of all an asymptote is a line that is close to an axis but never touches it. Now look at the graph as a normal reciprocal graph of y=1/x th...
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Answered by Aniqah B. • Maths tutor
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If y=3x^3e^x; find dy/dx?

Using the product rule we know that dy/dx = uv' + vu' where u = 3x^3; v = e^x. e^x differentiates to itself multiplied by any number in front of the x. u' = 9x^2; v' = e^x. Therefore dy/dx = 3x^3e^x + 9x^2e^...
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Answered by Aimee G. • Maths tutor
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Why do you not add the 'plus c' when finding the area under a graph using integration even though you add it when normally integrating?

when integrating with the limits, you substitute in the two limits into the integral and then take them away. As both will have added the constant 'c', you are taking c away from c and so cancel out, so it i...
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Answered by Neha M. • Maths tutor
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For a given function F(x), what does the graph of the function F(x+2) look like in comparrison to F(x)?

F(x+2) is simply F(x) but whenever you see an 'x' replace it with 'x+2'. So when x is say 3, F(x) is F(3) and F(x+2) is F(5). If we draw this out on a graph, we see that this has the effect of shifting the g...
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Answered by James S. • Maths tutor
3884 Views

Find the x-values of the turning points on the graph, y=(3-x)(x^2-2)

The minimum point occurs where dy/dx=0 We have 2 options: 1.) Expanding the brackets 2.) The product rule of differentiation The shortest is the product rule: dy/dx= (d/dx)(3-x).(x 2 -2) + (3-x).(d/dx)(x 2 -...
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Answered by Zita E. • Maths tutor
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