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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.

KittyL
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Ella
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  • You have the right tangents. Can you figure out what angle each of the tangents makes to the $x$-axis? – Arthur Aug 11 '15 at 09:33
  • use the formula $\tan(\theta)=|\frac{m_1-m_2}{1+m_1m_2}|$ – Dr. Sonnhard Graubner Aug 11 '15 at 09:34
  • Arthur. I figure I will need rise over run which is basically the opposite over the adjacent, and that equals the gradient. So would it be correct if I use tan(-11) and tan(11) to find the angle each of the tangents makes with the x-axis? – Ella Aug 11 '15 at 09:46

3 Answers3

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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, $$\theta=\tan^{-1}\left(\frac{11}{60}\right)$$ $$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{acute angle:}\ \theta=\tan^{-1}\left(\frac{11}{60}\right)\approx10.39^\circ}}$$

$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{obtuse angle:}\ \theta=\pi-\tan^{-1}\left(\frac{11}{60}\right)\approx 169.61^\circ}}$$

Harish Chandra Rajpoot
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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 $\tan\beta=-11$,…

egreg
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  • Could you possibly give me a visual? I do not see how the trig identity can find the angle. – Ella Aug 11 '15 at 09:56
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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

Khaled
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