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secondary 3 | A Maths
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Shan Kuek
Shan Kuek

secondary 3 chevron_right A Maths chevron_right Singapore

Please help me out!!

Date Posted: 4 years ago
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Eric Nicholas K
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Here. Apparently I have lots of comments for this question, but the system somehow does not let my answer in when I type in too much comments.
Eric Nicholas K
Eric Nicholas K
4 years ago
Here.

Apparently I didn't expect linear law to appear so quickly during Sec 3 A Maths. I expected the topic to appear only in around July or August of this year.

The first thing we need to do is to convert the equation into a linear form. The given form is obviously not linear, as x is the power of an unknown whereas linear forms require both y and x to be not powers.

We have to use logarithmic approaches to bring the x down, since the power rule of logarithms are capable of bringing powers down. I use lg in this case (ln works perfectly fine too!)
Eric Nicholas K
Eric Nicholas K
4 years ago
A further touch up of the equation gets you lg y = (3 lg q) x + 3 lg p. Recognise that there are three terms here, namely lg y, (3 lg q) x and 3 lg p. We see that exactly two of the three terms contain x and/or y (lg y counts as well since x and y are data that we collect and we are able to compute lg y from our data). Also, the side which contains the "alone" lg y has no coefficients attached to it, which is typical of a straight line.

Therefore, this equation lg y = (3 lg q) x + 3 lg p is in the form of Y = mX + c where Y = lg y, m = 3 lg q, X = x and c = 3 lg p.

Now that we have obtained this linear form, we need to tabulate values of x and lg y from our values of x and y. We do not need to do anything to the x values, but we need to compute out the lg y values. From there, we plot the coordinates of (x, lg y) and plot them all out.
Eric Nicholas K
Eric Nicholas K
4 years ago
It is a requirement that the straight line which you have drawn must pass the vertical axis. This is because the intercept must be read out directly from there (rather than obtaining the equation of the line to find the intercept).

I need to draw x from 0 to 6 and there are 10 large squares of 2 cm each, so I use 2 cm to represent 1 unit on the horizontal x axis.

For every 1 unit change in x, I see that lg y changes by approximately 0.3. The lowest point is 0.06, which is around 0. I predict that the y-intercept will be somewhere around 1.8, so as a standby, I round this up to 2.

I need to draw lg y from 0 to 2 and there are 12 to 13 large squares of 2 cm each (depending on the graph paper used, some have 12, some have 13) so I use 2 cm to represent 0.2 units on the vertical lg y axis.
Eric Nicholas K
Eric Nicholas K
4 years ago
Now I draw the best fit line which is the "average" line which fits the data points best. Note that in a minority of circumstances, the best line may not even pass through any data point. Remember to extend the line to the lg y axis.

Once done, we find the gradient and read off the vertical intercept of the curve and equate it to our gradient (3 lg q) and intercept (3 lg p) to compute estimated values for p and q from our calculator.

For the last part, we need to obtain working from the graph, so with the given value y = 10, we need to convert this into lg y = 1 first before drawing the relevant working line.