Can anyone help find the minimum or turning point of y equals x2 plus 19x divided by x minus 1?

Answer 1

To find a stationary point such as a minimum, you need to find where the gradient, dy/dx = 0.

y = (x2 + 19x)/(x-1)

Because this has two parts, one divided by the other, the quotient rule should be used to differentiate.

Quotient rule: If y = u/v, then dy/dx = (v*(du/dx) - u*(dv/dx))/v2

So in this case dy/dx = ((x-1)(2x+19) - 1(x2+19x))/(x-1)2

= (2x2 + 17x -19 - x2 - 19x)/(x-1)2

= (x2 - 2x - 19)/(x-1)2

For this to be 0, x2 - 2x - 19 must be 0.

(x-1)2 - 20 = 0

(x-1)2 = 20

x-1 = +- sqrt 20

x = 1+2sqrt5, 1-2sqrt5

Substituting these values into the original equation will find the y coordinate of the points.

x = 1+2sqrt5

y = (21 + 4sqrt5 + 19 + 38sqrt5)/2sqrt5

y = 21 + 4sqrt5

x = 1-2sqrt5

y = (21 - 4sqrt5 + 19 - 38sqrt5)/-2sqrt5

y = 21 - 4sqrt5