3

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my question is how can i get a equation out of this i have gathered the following

$$ y = a|x-h| + k$$

$$y = a|x-1| - 4$$ ($a$ i don't know yet and this is where i have a question)

to find $a$ can i just use any point in the graph and plug it in the equation or is their any other ways of doing it?

lulu
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MATH ASKER
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2 Answers2

2

This is $$y = a|x-1| - 4$$ your equation. It's a translation of the modulus function by 1 unit in the positive $x$ direction and by 4 units in the negative $y$ direction. The gradient is given by $a$.

You can see it's in the form $y=mx+c$ with $m=a$, $c=-4$ and $x=\mid x-1 \mid$ but this is the equation for a straight line. However, the modulus takes two values, namely, $x-1$ and $1-x$, so there will be 2 straight line equations.

When dealing with modulus functions you must consider the two separate cases to find the component lines.

So the equation of the left line is $\color{blue}{y=a(1-x) -4}$ and the right line has equation $\color{red}{y=a(x-1) -4}$.

$a$ can be found by substituting in a point that lies on the left line; for example substitute $(-1,0)$ into the $\color{blue}{\mathrm{blue}}$ equation and solve for $a$:

$$\color{blue}{0=a(1--1) -4}$$ $$\implies \color{blue}{0=2a -4}$$ $$\implies \color{blue}{a=2}$$

Substituting $(3,0)$ into the $\color{red}{\mathrm{red}}$ equation gives: $$\color{red}{0=a(3-1) -4}$$ $$\implies \color{red}{0=2a -4}$$ $$\implies \color{red}{a=2}$$

as before.

BLAZE
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  • so to find A i can use any point of the graph as long as the graph has gone trough it right? – MATH ASKER Oct 21 '15 at 23:17
  • @MATHASKER Correct, but choose the appropriate component equation – BLAZE Oct 21 '15 at 23:17
  • so the appropriate equation would be a(x-1)-4 right? and i have another question does this apply to any if given graph and asked to find the equation...if it is greatest equation graph? – MATH ASKER Oct 21 '15 at 23:21
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    @MATHASKER That's right, as long as the graph passes through that point you can substitute it in and solve it, I will show you that you can get the same answer for $a$ using the $\color{red}{\mathrm{red}}$ equation as well. – BLAZE Oct 21 '15 at 23:24
  • @MATHASKER See my answer please, I have edited it to show you it can be done with either line (red or blue) – BLAZE Oct 21 '15 at 23:29
  • Ohh i see now thanks....I have an exam tommorow and i was thinking can i just plug in the point in lets say the first equation, the one with |x| in it. and if i get a negative inside these box || it means i have to turn it to positive right? will that effect the A? – MATH ASKER Oct 21 '15 at 23:32
  • @MATHASKER My best advice is separate them out into component equations first before you go substituting anything in (like i did here) and good luck with exam tomorrow! – BLAZE Oct 21 '15 at 23:35
  • @MATHASKER Is there anything else? – BLAZE Oct 21 '15 at 23:40
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    Thanks...and if i don't get the same a in both of the equation than that means something is wrong right? – MATH ASKER Oct 21 '15 at 23:40
  • @MATHASKER That's right, it will mean something has gone wrong. – BLAZE Oct 21 '15 at 23:42
1

It seems that in your equation $a=2$ to make the equation $y=2|x−1|−4$.

Verify this from the points: $(1, -4)$, $(3,0)$, and $(-1, 0)$.

$a$ is the slope of your equation.

Sirius Black
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  • how did you find A? – MATH ASKER Oct 21 '15 at 23:15
  • I took a few points where the x and y intersect and was able to guess. –  Oct 21 '15 at 23:18
  • The proper way to do it is to use the formula m = (y0 - y1)/(x0-x1) –  Oct 21 '15 at 23:19
  • For example, if you take (x0, y0) = (3,0) and (x0, y0) = (1,-4), then plugging in, m = (0-(-4))/(3-1) = 2 –  Oct 21 '15 at 23:21
  • I would avoid taking points from different slopes. For example, don't use (-1, 0) and (1, -4) to find m because their slopes obviously not equal. –  Oct 21 '15 at 23:23
  • y0 x0? are they the x and y intercepts from each side of the graph? – MATH ASKER Oct 21 '15 at 23:24
  • Nah, it's just a point. Think of the first point having zero subscripts and the second point having 1 subscripts. It's the "slope formula." –  Oct 21 '15 at 23:27
  • ohh yes i remember the slope formula thanks a lot for your answer... – MATH ASKER Oct 21 '15 at 23:28
  • Sorry for the confusion above. I meant (x0, y0) and (x1, y1), not the subscripts both being 0. –  Oct 21 '15 at 23:30