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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: naved_s9994 on October 10, 2009, 03:57:43 pm

Title: Cos Graphh
Post by: naved_s9994 on October 10, 2009, 03:57:43 pm: The method I use to sketch the graph that follows (for NON-CAS), is seeming to be a
long method, which is quite time consuming and now a concern, given that we only get 1 hour in the real exam.

How would YOU go about sketching this graph from f:[-pi, pi] --> R, f(x)= 5cos(2(x+pi/3))
How would you find the intercepts at which it crosses the x-axis.

Any techniques?

Thank-you :)
Title: Re: Cos Graphh
Post by: TrueTears on October 10, 2009, 04:07:22 pm: Consider the graph of $y = cos(x)$

for $x \in [0, 2\pi]$

we have several important points, namely:

$\{0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi \}$

Now let $X = 2 \left(x+\frac{\pi}{3}\right)$

$x = \frac{X}{2} - \frac{\pi}{3}$

Thus for the graph of $y = f(x)$ the 'important' points will be at

$\{\frac{0}{2} - \frac{\pi}{3}, \frac{\frac{\pi}{2}}{2} - \frac{\pi}{3}, \frac{\pi}{2} - \frac{\pi}{3}, \frac{\frac{3\pi}{2}}{2} - \frac{\pi}{3}, \frac{2\pi}{2} - \frac{\pi}{3} \}$

Mark these in as a dot on the x axis.

Now the corresponding y values are:

$\{5, 0, -5, 0, 5, \mbox{etc}\}$

Find the y axis intercept, ie, $f(0)$ .

Now we can see the difference in the 'important' points is $\frac{\pi}{4}$ , so $\pm \frac{\pi}{4}$ to the 'important' points until you have all the points within the domain $[-\pi, \pi]$ and then put in corresponding y value. Then trace over the dots and you are done!

NOTE: vertical translation/dilation won't affect these "important" points.
Title: Re: Cos Graphh
Post by: bem9 on October 10, 2009, 04:08:25 pm: get the y intercept at (0,-2.5) and make it the endpoints too (pi, -2.5) & (-pi, -2.5)

next take the point 0,5 and translate it across by pi/3, this is a maximum
add or minus the period to that point until out of the domain to get other maximums

go half way between two maxiumums and there is a minimum at y=-5, also add or minus the period to this to get other minimums

for x ints, they occur halfway between a max and a min, so use that method to find them

join the dots!
Title: Re: Cos Graphh
Post by: kdgamz on October 10, 2009, 04:11:10 pm: FInd the period
find the amplitude
sketch it lightly with a pencil
then translate all x intercepts and stationary points accordingly....
to find the end-points just substitute the max and min values of the domain interval
finally find the y intercept

thats how i do it
Title: Re: Cos Graphh
Post by: naved_s9994 on October 10, 2009, 04:16:55 pm: Quote from: TrueTears on October 10, 2009, 04:07:22 pm
Consider the graph of $y = cos(x)$

for $x \in [0, 2\pi]$

we have several important points, namely:

$\{0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi \}$

Now let $X = 2 \left(x+\frac{\pi}{3}\right)$

$x = \frac{X}{2} - \frac{\pi}{3}$

Thus for the graph of $y = f(x)$ the 'important' points will be at

$\{\frac{0}{2} - \frac{\pi}{3}, \frac{\frac{\pi}{2}}{2} - \frac{\pi}{3}, \frac{\pi}{2} - \frac{\pi}{3}, \frac{\frac{3\pi}{2}}{2} - \frac{\pi}{3}, \frac{2\pi}{2} - \frac{\pi}{3} \}$

Mark these in as a dot on the x axis.

Now the corresponding y values are:

$\{5, 0, -5, 0, 5, \mbox{etc}\}$

Find the y axis intercept, ie, $f(0)$ .

Now we can see the difference in the 'important' points is $\frac{\pi}{4}$ , so $\pm \frac{\pi}{4}$ to the 'important' points until you have all the points within the domain $[-\pi, \pi]$ and then put in corresponding y value. Then trace over the dots and you are done!

NOTE: vertical translation/dilation won't affect these "important" points.

Thanks so much!
Also, can this technique be applied upon sin graphs?
Title: Re: Cos Graphh
Post by: TrueTears on October 10, 2009, 04:17:45 pm: Quote from: naved_s9994 on October 10, 2009, 04:16:55 pm
Quote from: TrueTears on October 10, 2009, 04:07:22 pm
Consider the graph of $y = cos(x)$

for $x \in [0, 2\pi]$

we have several important points, namely:

$\{0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi \}$

Now let $X = 2 \left(x+\frac{\pi}{3}\right)$

$x = \frac{X}{2} - \frac{\pi}{3}$

Thus for the graph of $y = f(x)$ the 'important' points will be at

$\{\frac{0}{2} - \frac{\pi}{3}, \frac{\frac{\pi}{2}}{2} - \frac{\pi}{3}, \frac{\pi}{2} - \frac{\pi}{3}, \frac{\frac{3\pi}{2}}{2} - \frac{\pi}{3}, \frac{2\pi}{2} - \frac{\pi}{3} \}$

Mark these in as a dot on the x axis.

Now the corresponding y values are:

$\{5, 0, -5, 0, 5, \mbox{etc}\}$

Find the y axis intercept, ie, $f(0)$ .

Now we can see the difference in the 'important' points is $\frac{\pi}{4}$ , so $\pm \frac{\pi}{4}$ to the 'important' points until you have all the points within the domain $[-\pi, \pi]$ and then put in corresponding y value. Then trace over the dots and you are done!

NOTE: vertical translation/dilation won't affect these "important" points.

Thanks so much!
Also, can this technique be applied upon sin graphs?
Sure can.