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The curve C is defined by x^3 – (4x^2 )y = 2y^3 – 3x – 2. Find the value of dy/dx at the point (3, 1).

When we find dy/dx we find the gradient of the curve at (3,1). Start by differentiating the left hand side (LHS) like so.. (whiteboard). Remember every time we differentiate a y value we multiply by dy/dx. T...
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Answered by Caitlin R. • Maths tutor
5550 Views

Sketch the curve y=4-(x+3)^2, showing the points where the curve crosses the x-axis and any minimum or maximum points.

This equation rearranges to give -y=(x+3)^2-4, which is very similar to our curve y=(x+3)^2-4 from before. In fact, replacing y with -y in an equation is equivalent to reflecting the curve through the x-axis...
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Answered by Jonny I. • Maths tutor
4282 Views

Find the coordinates of the minimum point of the curve y=x^2+6x+5.

To answer this question is equivalent to minimising y=(x+3)^2-4. We have that all square numbers are greater than or equal to 0 so to minimise this equation, we require that (x+3)^2=0. This is satisfied only...
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Answered by Jonny I. • Maths tutor
13030 Views

Factorise x^2+6x+5=0 by completing the square.

When completing the square, we first divide the whole equation by the x^2 component. In this case, the x^2 component is 1 so nothing changes. We now apply the method to convert to square form: we reduce the ...
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Answered by Jonny I. • Maths tutor
10886 Views

Find the coordinates where the curve y=x^2+6x+5 crosses the x-axis.

When any curve crosses the x-axis, the y-coordinate is 0 at that point. Hence, our answers will have y=0. So we want to solve x^2+6x+5=0. From before, we have that x^2+6x+5=0 can be rewritten as (x+5)(x+1)=0...
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Answered by Jonny I. • Maths tutor
5184 Views