Tangents and Normals at a given point

The derivative is given by $\frac{dy}{dx}=2x$ and so the gradient when $x=3$ is $m=6$. So far, we have that the equation of the tangent is $y=6x+c$. To find $c$, we use the fact that the point $(3,9)$ is on the tangent as well as the curve (it is the only point) and so when $x$ is 3, we should get $y=9$: $9=6\times 3+c$. This gives $c=-9$ and the equation of the tangent is therefore $y=6x-9$.

The derivative is given by $\frac{dy}{dx}=6x^2+2x-3$. At $x=-0.5$, $dy/dx=-2.5$. This is the gradient of the curve and hence the gradient of the tangent at $x=-0.5$. The tangent will have equation $y=-\frac{5}{2}x+c$, where we need to find the $y$-coordinate in order to find $c$. The $y$-coordinate at $x=-0.5$ is $y=-5.5$ and so $-\frac{11}{2}=-\frac{5}{2}\times -\frac{1}{2}+c$ giving $c=-\frac{27}{4}$. It follows that the equation of the tangent is $y=-\frac{5}{2}x-\frac{27}{4}$ or rearranging into the required form: $10x+4y+27=0$.

We saw in the above example that the gradient of the tangent is 6, and so the gradient of the normal is $m=-1/6$. So far we have that the equation of the normal is $y=-\frac{1}{6}x+c$. To find $c$, we use the fact that the point $(3,9)$ is on the normal as well as the curve (the only point that the curve, tangent and normal have in common) and so when $x$ is 3, we should get $y=9$: $9=-\frac{1}{6}\times 3+c$. This gives $c=9.5$ or $19/2$ and the equation of the tangent is therefore $y=-\frac{1}{6}x+\frac{19}{2}$.

The line $2y+x=5$ can be rewritten as $y=-\frac{1}{2}x+\frac{5}{2}$. If the normals are parallel to this line, it means that the normals have gradient $-1/2$. It follows that the tangents and the curve at points $A$ and $B$ all have gradient 2. Hence, we must solve $f'(x)=2$:

$\frac{x}{3}-\frac{2}{x}+\frac{7}{3}=2$ i.e. $\frac{1}{3}x^2-2+\frac{7}{3}x=2x$

or $x^2+x-6=(x-2)(x+3)=0$. It follows that the $x$-coordinates of $A$ and $B$ are $x=2$ and $x=-3$.