This is quite graphic...

Simplificatation, evaluation, linear equations, linear graphs, linear inequalities, basic word problems, etc.
diaste
Posts: 14
Joined: Sat Jan 17, 2009 1:54 pm
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This is quite graphic...

I don't know why I'm having so much trouble with this, obviously I'm missing something.
SO here goes.
I have a tent that's going to be staked out on a polar ice cap and as you can imagine it's cold and windy out here. My partner is the meticulous type and wants to know how high the tent is so he can put the stakes just where the instructions say. I think that's a bit overboard as I'm cold and I want to make some tea but I relent and tell him I can figure it out for him. The only info I get from the instructions is that the tent lies on the graph of: $y = - 2|x-5| + 10$ and the tent ropes are supposed to lie on the graph of: $y = - \frac13 |x-5| + 6$ From the first equation I quickly assess the height of the tent to be 10' and the base to be ten feet with the slope of $- \frac21$ thrown in for good measure. This impresses my partner but it's the cold that's motivating me. However, now we're in trouble because I told him the tent stakes should go 18' from the edge of the tent and we don't have that much rope.

Seriously, this equation $- \frac13 |x-5| + 6$ is baffling me. The vertex is (5,6) and the slope is $- \frac13$ but I keep coming up with x intercepts at -23 and 23. This is what I did:
$-\frac 13|x-5| + 6$
$-\frac 13|x-5|= -6$
$-\frac 13x+1.666 = -6$
$-\frac 13x = -7.666$
$x = 23$ Of course that's wrong because the graph demands a slope of -1 to 3 and that means I should have 18 and -18 as an answer. Which, from the vertex would put the stakes at 13' from the edge of the tent.

I thought of just eliminating the absolute value and dividing 6 by $\frac 13$ because that would work. Is that the way I should treat a negative slope that's greater than 1? The first equation worked out just fine with intercepts at ( 0,0) and (10,0) and the vertex at (5,10) I assume there has to be a set of rules to follow I just can't seem to find what they are. I'm going to try one more idea I just thought of and then come back and look for a reply.

One thing I would request is that you remember I'm a mere math mortal and a pretty green one at that. You two are way advanced!

Thanks for your patience and please hurry, it's winter in the arctic regions ya know!

stapel_eliz
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Joined: Mon Dec 08, 2008 4:22 pm
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diaste wrote:Seriously, this equation $- \frac13 |x-5| + 6$ is baffling me. The vertex is (5,6) and the slope is $- \frac13$ but I keep coming up with x intercepts at -23 and 23. This is what I did:
$-\frac 13|x-5| + 6$
$-\frac 13|x-5|= -6$
$-\frac 13x+1.666 = -6$

Absolute-value bars are not the same as parentheses. You have to deal with cases, accounting for the possible sign of the argument. For absolute-value equations, that means taking the bars off twice: one way positive and the other way negative:

. . . . .-(1/3)|x - 5| + 6 = 0

. . . . .-(1/3)|x - 5| = -6

. . . . .(1/3)|x - 5| = 6

. . . . .(1/3)(x - 5) = 6 or (1/3)(x - 5) = -6

. . . . .x - 5 = 18 or x - 5 = -18

...and so forth.

It should be noted that the stakes' rope is measured from the ends of the tent, not from the center.

Eliz.

diaste
Posts: 14
Joined: Sat Jan 17, 2009 1:54 pm
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