Question

y=px-ap(1-p) solve it

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Answer 1

Answer:

y=px-ap(1-p) (Given question)

y=px -ap+ap^2

y=p(x-a+ap)

Using y=mx+c intercept-form we get:-

y=p(x-a+ap)

y=p{x+a(p-1)}

Hence,y=p{x+a(p-1)} where the value of c is constant.

Answer 2

The value of the constant c is  $\frac{x-a}{2a}$

Given :

y = px - ap(1-p)

To Find:

To get the value of y for the equation px - ap(1-p)

Solution:

y = px - ap(1-p)

It can be written as ,

= p(x-a) + a$p^{2}$............(1)

Differentiating partially with respect to 'p',

We have,

0 = x - ap(-1) - (1-p)a

0 = x + ap -a + ap

p = $\frac{x-a}{2a}$...................(2)

Substituting, the value of p in equation (1),

y = (x-a)$(\frac{x-a}{2a})$  + a $(x-a)^{2}$

y = $\frac{(x-a)^{2} }{2a} (1-\frac{1}{2} )$

y = $\frac{(x-a)^{2} }{4a}$.............(3)

General solution can be written as,

y = cx +ac (1-c)

c = $\frac{x-a}{2a}$ which is the required answer

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