The-graph-of-y-a-bx-x-1-x-4-has-a-turning-point-at-P-2-1-Find-the-value-of-a-and-b-and-hence-sketch-the-curve-y-f-x-showing-clearly-the-turning-points-asympotote

Question Number 83964 by Rio Michael last updated on 08/Mar/20

The graph of

y = \frac{a + b x}{(x - 1) (x - 4)}

has a turning point at P (2, - 1) . Find the value of a and b

and hence, sketch the curve y = f (x) showing clearly the

turning points, asympototes and intercept (s) with the

axes .

Answered by john santu last updated on 08/Mar/20

(1) - 1 = \frac{a + 2 b}{1 . (- 2)} \Rightarrow a + 2 b = 2

(2) y = \frac{2 - 2 b + b x}{x^{2} - 5 x + 4}

y^{'} = \frac{b (x^{2} - 5 x + 4) - (2 x - 5) (2 - 2 b + b x)}{{(x - 1)}^{2} {(x - 4)}^{2}}

y^{'} = 0 for x = 2

- 2 b - (- 1) (2) = 0

2 b = 2 \Rightarrow b = 1 \land a = 0

∴ y = \frac{x}{x^{2} - 5 x + 4} .

v e r t i c a l a s y m t o t e x = 1 \land x = 4

h o r i z o n t a l a s y m t o t e

y = \lim_{x \to \infty} \frac{x}{x^{2} - 5 x + 4} = 0

Commented by jagoll last updated on 08/Mar/20

Commented by jagoll last updated on 08/Mar/20

this is the graph y = \frac{x}{x^{2} - 5 x + 4}

Commented by Rio Michael last updated on 08/Mar/20

t h a n k s s i r s g r e a t w o r k

i a p p r e c i a t e