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I got a question from my textbook, I don't know how to solve the second question (Proof). Anyone can help me.

Find the equation of the tangent at the point $P(3, 9)$ to the curve $y = x^3 - 6x^2 + 15x -9$. If $O$ is the origin, and $N$ is the foot of the perpendicular from $P$ to the $x$-axis, prove that the tangent at $P$ passes through the mid-point of $ON$.

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2 Answers 2

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Hint:

  • determine $y'$
  • determine the equation of the tangent $t$
  • find the zero of the tangent $t$
  • the perpendicular from $P$ to the $x$-axis shares the $x$-coordinate with point $P$
  • point $N$ is on the perpendicular, with $y$-coordinate $0$

enter image description here (Large version)

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First, you must compute the derivative of y(x), which is $y'(x)=3x^2-12x+15$, and so $y'(3)=6$, a value that will be the slope of the tangent line to $y(x)$ at the point $P=(3,9)$. Now we use the following expression for a line, where $(x_0,y_0)$ is a point of the line (in this case $P=(3,9)$) and $m$ is the slope at that point (in this case $m=6$):

$y-y_0=m(x-x_0)$

we conclude that the equation of the tangent line we are looking for is $y-9=6(x-3)$, that is, $y=6x-9$. We must see that, if $N$ is the foot of the perpendicular from $P$ to the $x$-axis, then the previous tangent line passes through the mid-point of $ON$, so let's calculate the explicit expression of these elements. The foot will be obviously the point $(3,0)$, and since the point coordinates of the origin are $(0,0)$, the midpoint of $ON$ will be $Q=(3/2,0)$.

And does $Q$ belong to the tangent line? Yes, because $6\cdot(3/2)-9=0$, thus completing the proof.