1. ## Calc 1 problem

hey, can anybody help solve this HW question? thanks!
Write a function that models the distance D from a point on the line y = 3 x - 6 to the point (0,0) (as a function of x).

2. Originally Posted by smoti10
hey, can anybody help solve this HW question? thanks!
Write a function that models the distance D from a point on the line y = 3 x - 6 to the point (0,0) (as a function of x).

A point in the plane belongs to your line iff it is of the form $(x,3x-6)$ . Well, now write down the formula for the distance from this generic

point on the line to the origin (0,0).

Tonio

3. Each point on the line can be expressed as a coordinate in the form...

$(x, 3x - 6)$.

Why? A coordinate is of the form $(x,y)$ and $y = 3x - 6$

So what to do is think in terms of Pythagoras triangle... You have two sides, one of length $x$, one of length $3x-6$ and one unknown side that you must find.

Why? Well just think of the coordinate $(x, 3x - 6)$ as being distance along the x-axis then distance up to get to the point. This will form a right angled triangle with side lengths $x$ and $3x -6$

So, let's call this unknown side $d$, using the standard Pythagoras formula we get...

$d^2 = x^2 + (3x - 6)^2.$

Now you finish this off by expanding everything and you will be left with a formula of the form...

$d^2 = ...$

then just take square roots to get,

$d = \sqrt{\dots}$

4. ## thanks for your help

thank you very much for your help! now,once you got sqrt(10x^2 - 36x + 36) how can you find out where is the closest point to the point (0,0) on that line?
Thanks again.

5. Originally Posted by smoti10
thank you very much for your help! now,once you got sqrt(10x^2 - 36x + 36) how can you find out where is the closest point to the point (0,0) on that line?
Thanks again.
Well you want to find the value of $x$ that makes that function the smallest.

This will happen at one of two places.

A critical point,
An end point of the range of $x$. (i.e. $x = \pm \infty$).

Immediately you can see that that if $x \to \pm \infty$ then $f(x) \to \infty$ (if not I'll explain).

So that leaves you to find critical points...

I'll start you off...

If $f(x) = \sqrt{10x^2 - 36x + 36}$

=> $f'(x) = \frac{2(5x-9)}{\sqrt{10x^2 - 36x + 36}}$.

Solve for this to be 0 and what does that tell you..?

6. Not "fixed point", critical point. A fixed point for a function, f(x), is a value of x such that f(x)= x.

7. Originally Posted by HallsofIvy
Not "fixed point", critical point. A fixed point for a function, f(x), is a value of x such that f(x)= x.
Oops! Yes correct, I'm currently working away on my dynamical systems tutorials so probably just got a bit of cross subject banter lol.