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dy/dx at (0,2) = 0
i don't know what to do next.

@ startanewww what did you get on your derivative at (0,2)?

isn't it 0?

after getting the slope at (0,2) find the normal means perpendicular to the line at (0,2)

1st find the equation of tangent to the curve...

slope perpendicular to it is m2= - 1/m1

y-y1 = m(x-x1)

but the slope I found for dy/dx at (0,2) is zero. What can i do?

then, the slope of the normal will be -1/0 = infinity
how can i find the equation of normal?

Diffrentiate y=(x+2)e^x-3x

u know Product Rule ?

u get slope = -1

SO...Slope of Normal = -1 / m1 = -1/-1 = 1

Nw just write the equation using points (0,2) m = 1
y-y1 = m(x-x1)

y'=e^x+(x+2)e^x-3
dy/dx at (0,2) = e^0 + (0+2)e^0 - 3 = 0

am i correct?

y=(x+2)e^x-3x
y' = (x+2) *1 + e^x *1 - 3

differentiating e^x gives e^x right?

am i correct?

lets say y'=0, can you form a line perpendicular to the curve? try it :D

first fint an equation of the line tangent to the curve at (0,2)

(y-2)/(x-0)=0
y=2?

what about the equation of normal?

yes correct :D then ?

what can i do?

the slope will be infinity....

what is a line perpendicular to line y=2?

at (0,2) ?

will the slope of the normal be infinity ?

|dw:1356675309579:dw|

but -1/0 = infinity?

x=0 is the line

i don't know the answer....

i think you are right

ah ok theres no answer in the book? :D

my classmates said the answer was x=0

hmm the line perpendicular to y=2 is the y axis or the vertical line

ok so the answer should be x=0 only?

hmm it can be x=0

x = 0 to infinity?

it could be y= 0 to 2, or 2 to infinity, or y= 0 to negative infinity

?????? how come

equation of the normal at (0,2)...? should there be only one answer?

vertical lines

the vertical line y axis

yess correct :D

so what is the final answer? x=0 ?

i would say at point (0,2) it is y= from 0 to 2 vertical line

ok thanks

ok YW,,, good luck now :D