+118658 \(y=\frac{(x^2+4)(x-3)}{2x}\qquad x\ne0\)
y=u/v u=(x^2+4)(x-3) v= 2x
u'= 2x(x-3)+1(x^2+4) v'=2
u' = 3x^2 -6x +4
\(y=\frac{u}{v}\\ y'=\frac{vu'-uv'}{v^2}\\ y'=\frac{(2x)(3x^2-6x+4)-(x^2+4)(x-3)*2}{(2x)^2}\\ y'=\frac{\not{2}(3x^3-6x^2+4x)-(x^2+4)(x-3)*\not{2}}{\not{4}^{\;2}x^2}\\ y'=\frac{(3x^3-6x^2+4x)-(x^2+4)(x-3)}{2x^2}\\ y'=\frac{(3x^3-6x^2+4x)-(x^3-3x^2+4x-12)}{2x^2}\\ y'=\frac{3x^3-6x^2+4x-x^3+3x^2-4x+12}{2x^2}\\ \frac{dy}{dx}=\frac{2x^3-3x^2+12}{2x^2}\\\)
\(y=\frac{(x^2+4)(x-3)}{2x}\\ when\;\; x=-1\\ y=\frac{(1+4)(-1-3)}{-2}\\ y=\frac{(1+4)(-1-3)}{-2}\\ y=\frac{(5)(-4)}{-2}\\ y=10\\ (-1,10)\\~\\ \)
\(y'(-1) = \frac{-2-3+12}{2}=\frac{7}{2}\quad \text{This is the gradient of the tangent at x=-1}\)
To find the equation of the tangent you need to simplify this
\(\frac{y-10}{x--1}=\frac{7}{2}\)
Help if possible please :)
