Can sum 1 plz tell me how to do these question.

Thanks very much.

## 2003 Paper 1 Q11 (b) & Q12

### Allan

Posted 20 May 2004 - 05:27 PM

**for 11b**

the parabola crosses the a-axis at 0 and 24

so you can put:

y= kx(x-24)

subsitute in a point e.g. (12,-5) (point A)

-5 = 12k(12-24)

-5 = 12k(-12)

-144k = -5

k = 5/144

y = 5/144x(x-24)

Therefore p = 5/144 and q = -24

**Q12**

Rearrange the equation using log rules:

log[e](2e)^3 - log[e](3e)簡

Using log rules:

log[e][(2e)^3/(3e)簡]

log[e][8e^3/9e簡]

Cancelling out e簡 you get

log[e](8e/9)

Using log rules:

log[e]8e - log[e]9

log[e]8 + log[e]e - log[e]9

log[e]e = 1

So log[e]8 + 1 - log[e]9

Therefore in the form the question asks for:

1 + log[e]8 - log[e]9

### babeasc

Posted 14 May 2007 - 11:33 AM

**for 11b**

the parabola crosses the a-axis at 0 and 24

so you can put:

y= kx(x-24)

subsitute in a point e.g. (12,-5) (point A)

-5 = 12k(12-24)

-5 = 12k(-12)

-144k = -5

k = 5/144

y = 5/144x(x-24)

Therefore p = 5/144 and q = -24

I understand the steps, but I dont understand why you can make q=(-24)?

the middle of the graph has only been moved 12 places to the right from the origin??