Finding the equations of the tangents where a quadratic equation cuts the $x$-axis and the angle between the tangents (differentiation involved) Calculate the equations of the tangent where $y=x^2-5x-24$ cuts the $x$-axis. 
$(x-8)(x+3)$ factorising
$x=8,  x=-3 $
$y'(x)=2x-5$
$y'(8)=11$
$y'(-3)=-11$
$y=11x+c$
$0=11(8)+c$
And then I find, $c$, and repeat for the other tangent equation which gives:
$y=11x-88$ and $y=-11x-33$
The second question is what is the angle between the tangents and I don't know how to find it. I know it has something to do with tan-theta.
Could someone also check if my arithmetic is correct for the first part.
 A: Notice, the slope of tangent at general point of the curve $y=x^2-5x-24$ $$\frac{dy}{dx}=\frac{d}{dx}(x^2-5x-24)=2x-5$$ Now, the point where curve  $y=x^2-5x-24$ cuts the x-axis has $y=0$ thus we have $$0=x^2-5x-24$$ Solving the quadratic equation for the values of $x$ as follows  $$x=\frac{-(-5)\pm\sqrt{(-5)^2-4(1)(-24)}}{2(1)}$$ $$=\frac{5\pm 11}{2} \iff x=8, -3$$ Hence, we get two points $(8, 0)$ & $(-3, 0)$ where curve intersects x-axis 
Now, the slope of the tangent at $(8, 0)$ is $$=\frac{dy}{dx}|_{x=8}=2\times 8-5=11$$ Hence its equation $$y-0=11(x-8)$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{y=11x-88}}$$
Similarly, the slope of the tangent at $(-3, 0)$ is $$=\frac{dy}{dx}|_{x=-3}=2\times (-3)-5=-11$$ Hence its equation $$y-0=-11(x-(-3))$$ 
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{y=-11x-33}}$$
Hence, the angle between the tangents is given as $$\tan \theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|$$ now, setting $m_1=11$ & $m_2=-11$, we get
$$\tan \theta=\left|\frac{11-(-11)}{1+11(-11)}\right|$$ 
$$=\left|\frac{11}{-60}\right|=\frac{11}{60}$$  Hence, $$\theta=\tan^{-1}\left(\frac{11}{60}\right)$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{acute angle:}\ \theta=\tan^{-1}\left(\frac{11}{60}\right)\approx10.39^\circ}}$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{obtuse angle:}\ \theta=\pi-\tan^{-1}\left(\frac{11}{60}\right)\approx 169.61^\circ}}$$
A: The first part is good.
The slope of a line is the (trigonometric) tangent of the angle the line forms with the positive $x$-semiaxis. If $\alpha$ is the angle formed by the tangent at $(8,0)$ and $\beta$ is the angle formed by the tangent at $(-3,0)$, you want to compute $\beta-\alpha$ and
$$
\tan(\beta-\alpha)=
\frac{\tan\beta-\tan\alpha}{1+\tan\beta\tan\alpha}
$$
Since $\tan\alpha=11$ and $\tan\beta=-11$,…
A: Basically the slope or the coeffecient of x in the tangent equation represents tan(theta)
So what you need to do is get artan(-11)= -84.81 then arctan(11)=84.81 and find the difference =169.62 deg you might want to take use the acute angel 10.39 deg
