Please show all work

label Calculus
account_circle Unassigned
schedule 1 Day
account_balance_wallet $5

Jul 2nd, 2015

Thank you for the opportunity to help you with your question! the solution is as such:

y= ab^x

for 1st point P  12= ab^ -2

for 2nd point Q 3/8 = ab ^3

on dividing the equations we get b= 1/2 and a = 3


now the tangent is parallel to PQ. so slope of tangent on the curve at the  required point is = slope of PQ. /

slope of PQ= (y2- y1) / (x2- x1)

                   = - 2.325

now slope of tangent on the curve at the  required point is -2.325

but we know slope of tangent is =dy/dx

so dy/dx =   d(ab^x)/dx 

              =  y lnb

now at (x,y)

the slope= -2.325

so y= 3.354

putting in the equation when y= 3.354 then x= 0.07

Please let me know if you need any clarification. I'm always happy to answer your questions.awaiting a positive feedback. It was lovely helping you. Looking forward to help you again. A positive feedback awaited thank you.
Jul 2nd, 2015

sorry x= -0.16

Jul 2nd, 2015

Studypool's Notebank makes it easy to buy and sell old notes, study guides, reviews, etc.
Click to visit
The Notebank
...
Jul 2nd, 2015
...
Jul 2nd, 2015
Sep 20th, 2017
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer