## Tuesday, 2 July 2013

### Linear Law worksheet Example 7 (By: Mason, Kai En, Shaun)

Firstly, we find the gradient and y-intercept, to be able to substitute in the values once we have made the equation in linear law form, Y = mX+c

y-intercept = 0.7 (as question states straight line cuts vertical axis at 0.70)

gradient = y2-y1/x2-x1 = 0.7-0 / 0-(-0.233) = 3.00 (3sf)

Now, we will make the equation y = 2-px^q into linear form.

px^q = 2-y
Apply lg to both sides
lg(px^q) = lg(2-y)
expand lg(px^q), and bring the power q to the front
lg(p) + qlg(x) = lg(2-y)

Now, we have an equation in the form Y=mX+c
lg(2-y) = qlg(x) + lg(p)
where lg(2-y) is Y
q is m,
lg(x) is X
and lg(p) is c

Hence, we can find the value of q by subbing in the gradient, and the value of lg(p) by subbing in the y-intercept.

q = 3

lg(p) = 0.7
p = 10^0.7
p = 5.01 (3sf)

Hence, p = 5.01, q=3

1. Very clear and easy to understand. The steps are systematic and neat. Good explanations, detailed.

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2. The explanation is very clear, especially when explaining how to get the linear form and which terms in the equation represent the Y, X, m and c.

3. The explanation is very clear, as it clearly and methodically explains how he gets to the linear form and which terms in the equation represent the Y, X, m and c.

4. The explanation is very clear, as it clearly and methodically explains how he gets to the linear form and which terms in the equation represent the Y, X, m and c.

5. The explanation of the workings is very clear and easy to understand.