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Firstly, an expression for dy/dx needs to be found to allow us to find the gradient of the normal. As a normal is a straight line, the equation y-y₁=m(x-x₁) can be used to find its e...

The first step would be to expand the second equation:(x-p)^2+qx^2-px-px+p^2+q

this simplifies to x^2-2px+p^2+q

After this you examine the two equations and identify their similariti...

4^x = (2*2)^x =2^x * 2^x = y^2

(2x+1)² < 9(x-2)²4x²+4x+1 < 9(x²-4x+4)4x²+4x+1 < 9x²-36x+360 < 5x²-40x+35Apply quadratic equation solving formula (...

In order to find the gradient of a tangent to the curve C we must differentiate our equation for C.dy/dx= 9x²-11To find the gradient of a tangent at a specific point P we substitute the coordin...