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${x^4}$ has two tangents at points P(x, y) and Q(x, y). These tangents both cross the point (-5/4, -8). Find the equations of these tangent lines.

Okay, so first I know that the equation for the slope from the derivative is ${4x^3}.$ and the equation for the other slope formula can be found by (y-y2)/(x-x2).

I know that the line passing through (-5/4, -8) is tangent to the curve at a point (a, b). I can use this to essentially make out $$ \dfrac{b + 8}{ a + 5/4} $$ and equate both formulas.

From here, I can single out $$b=4a^4 + 5a^3 - b - 8$$

Since the equation of the tangent lines is $y = m(x-x1) + y$, I would assume that I would substitute the slope $4a^3$ for m and then use a and b as points?

$$y=4a^3(x-a)+(4a^4 + 5a^3 - 8)$$ Am I on the right track here? I believe at this point my solution has become too complicated.

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You're right that the slope of the line through $(-\frac54,-8)$ and $(a,b)$ is $\frac{b + 8}{ a + 5/4}$, while the slope of the tangent line to $y=x^4$ at the point $(a,b)$ (assuming $b=a^4$, so that the point is on the curve) is $4a^3$. For these to be the same slope, we need $\frac{b + 8}{ a + 5/4} = 4a^3$, or $b+8=4a^4+5a^3$; and since we also need $b=a^4$, this becomes $a^4+8=4a^4+5a^3$. This equation, as it turns out, has exactly two real roots, which will give you the two $x$-coordinates you're looking for.

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  • $\begingroup$ Would these roots be 1 and -2? One equation of the tangent would be $y=-32(x+1)$ and the other would be $y=4x-3$ $\endgroup$ – user382540 Oct 28 '16 at 23:06
  • $\begingroup$ Almost perfect! The $x$-coordinates and slopes are good. Double-check the equation of the first line. $\endgroup$ – Greg Martin Oct 29 '16 at 3:02
  • $\begingroup$ Just out of curiosity, is there a way that you can factor this without a calculator?(I've become so used to the factoring capabilities on my calculator) $\endgroup$ – user382540 Oct 29 '16 at 13:30
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    $\begingroup$ We can always determine all the rational roots of a polynomial, by trying all rational numbers whose numerator divides the constant term and whose denominator divides the leading coefficient. Here that means trying numbers of the form $\pm1,\pm2,\pm4,\pm8$ as well as those numbers divided by $3$. In this case, we quickly find the roots $1$ and $-2$, and long dividing the polynomial by $a-1$ and $a+2$ yields a quadratic that (it's now easy to see) has no real roots. $\endgroup$ – Greg Martin Oct 29 '16 at 20:35
  • $\begingroup$ Okay, to be very honest I don't know if I understand but I tried out what you're saying $$ \dfrac{3a^4}{3a^4} + \dfrac{5a^3}{3a^4} - \dfrac{-8}{3a^3} $$ Now I just continue from here... $$1 + \dfrac{5}{3a} - \dfrac{-8}{3a^4} $$ And then I substitute rational values like you said(in this example maybe -2 $$ 1 + \dfrac{-5}{6} + \dfrac{-8}{48} $$ And I get... $$0=0$$ So does that essentially mean the rational value $(-2)$ I put in is a factor? $\endgroup$ – user382540 Oct 29 '16 at 21:25

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