# How do I solve that equation?

I have the function

$$f_t(x) = \frac{1}{4} x^4 + tx^3 - x^2$$

The assignment tells to check whether there are t for which the tangents on that functions inflextion points cross at $x = 0$.

My idea is to create the tangents and then check, whether their y-intercepts are equal. My problem is that by doing that there emerges a equation I am unable to solve.

The equation I get is the following:

$$(\alpha - t)^2\left[-3(\alpha - t)^2 - 8t(\alpha - t) + 4\right] = (\alpha +t)^2\left[5(\alpha + t)^2 - 16t(\alpha + t) - 12\right]$$

Where $\alpha = \sqrt{t^2 + \frac{2}{3}}$

I could expand that and see what happens though but I thought it would be a good idea to check for a different approach to learn something new.

If you need further information about how I achieved that equation just ask for it.

EDIT: I will attach the calculation of that equation so that my problem becomes clearer I hope:

The inflextion points are at $f''_t(x) = 0$ for which I get:

$$x_1 = -t + \sqrt{t^2 + \frac{2}{3}}$$ and

$$x_2 = -\left(t + \sqrt{t^2 + \frac{2}{3}}\right)$$

Let now $\sqrt{t^2 + \frac{2}{3}}$ be $\alpha$ and we get for the y-coordinate by applying $x_1$ and $x_2$ to $f_t(x)$:

$$y_1 = (\alpha-t)^2 \left(\frac{ (\alpha-t)^2}{4} + t (\alpha-t) - 1\right)$$

$$y_2 = (t+\alpha)^2 \left(\frac{ (t+\alpha)^2}{4} - t (t+\alpha) - 1\right)$$

Finally we get m by applying $x_{1/2}$ to $f'_t(x) = x^3 + 3tx^2 - 2x$:

$$m_1 = (\alpha-t) ( (\alpha-t)^2 + 3t (\alpha-t) - 2)$$

$$m_2 = -(t+\alpha) ( (t+\alpha)^2 - 3t (t+\alpha) - 2)$$

With $y = mx + b$ we get $b = y - mx$ for the y-intercept. We get its value by plugging in $x_{1/2}, y_{1/2}$ and $m_{1/2}$. Since I want to know about equality I would set $b_1$ equal to $b_2$ and solve for $t$. That then leads to the equation I am unable to solve.

Thank you very much for your help!

FunkyPeanut

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Why would that help? – FunkyPeanut Feb 20 '14 at 18:30
Ah, sorry, I thought you were looking for the roots. Let me delete my comment. – user99680 Feb 20 '14 at 18:30
What is an "apex" of a function? – Christian Blatter Feb 20 '14 at 19:27
That point where the function changes its direction. My translator also translates that into "inflextion point". You get its place on the x-Axis by setting $f''(x) = 0$. – FunkyPeanut Feb 20 '14 at 19:32