Hello, need help with this one question. Given that the curve y=ax^2+b/x has a gradient of -5 at the point (2,-2), find the values of a and b? Thanks

2 Answers
May 22, 2018

#a=-1 and b=4#.

Explanation:

The curve # C : y=ax^2+b/x# has a gradient #-5# at the point

#P(2,-2)#.

#:. [dy/dx]_(P(2,-2))=-5#.

#:. [d/dx{ax^2+b/x}]_(P(2,-2))=-5#.

#:.[2ax-b/x^2]_(P(2,-2))=-5#.

#:. 4a-b/4=-5......................................................(ast_1)#.

Further, #P in C#. Hence, the co-ordinates of #P# must satisfy the

eqn. of #C#.

#:. -2=4a+b/2, or, -4a-b/2=2......................(ast_2)#.

# (ast_1)+(ast_2) rArr -3b/4=-3 rArr b=4#.

#:.(ast_1)rArr 4a=b/4-5=4/4-5=-4," giving, "a=-1#.

May 22, 2018

#b=4#
#a=-1#

Explanation:

#y=ax^2+b/x#

#(dy)/(dx) = 2ax-b/x^2#

Since #(dy)/(dx)# can be used to find the gradient of the curve at the point #(2,-2)#, we can say:

#(dy)/(dx) = -5#

#2ax-b/x^2=-5#

And sub in #x=2#

#4a-b/4=-5# --- (1)

We can find the second equation by subbing in the point #(2,-2)# into the curve

#y=ax^2+b/x#

#-2=4a+b/2# --- (2)

From (1),

#4a-b/4=-5#

#16a-b=-20#

#b=16a+20# --- (3)

Sub (3) into (2)

#-2=4a+b/2#

#-2=4a+1/2(16a+20)#

#=-2=4a+1/2(2)(8a+10)#

#-2=4a+8a+10#

#12a=-12#

#a=-1# --- (4)

Sub (4) into (3)

#b=16a+20#

#b=16(-1)+20#

#b=-16+20#

#b=4#