anonymous
  • anonymous
Someone Please Help, Giving A Medal A.Find the slope of the tangent line to the graph of f at the given point. A.Find the slope-intercept equation of the tangent line to the graph of f at the given point.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
2 Attachments
welshfella
  • welshfella
find the derivative f(x) = 4x + 2 can you find f'(x)?
anonymous
  • anonymous
-1/2+x/4

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anonymous
  • anonymous
no the derivative of that is 4
welshfella
  • welshfella
yes thats right so the slope is - at every point on the graph as its a straight line
welshfella
  • welshfella
* slope is 4
welshfella
  • welshfella
the second function is:- f(x) = x^(1/2) can you find the derivative of this one?
anonymous
  • anonymous
x^-1/2/2
welshfella
  • welshfella
|dw:1435264790652:dw|
welshfella
  • welshfella
so to find the slope at (16,4) plug x = 16 into the the above value for f'(x)
welshfella
  • welshfella
then use the general form y-y1 = m(x-x1) and plug in slope m and the point (x1,y1) will be (16,4)
anonymous
  • anonymous
for the first step is it 2
welshfella
  • welshfella
1 / 2*x(1/2) = 1 / 2* (16)^1/2 = 1/ 2*4 = 1/8
welshfella
  • welshfella
so m = 1/8
anonymous
  • anonymous
im lost. I don't understand this at all.
welshfella
  • welshfella
m stands for the slope in the formula the slope is given by the derivative of sqrt x - and is 1 / 2* x (1/2) o r 1 / 2 sqrtx here x = 16 ( point is (16/4) so m = 1 / 2 sqrt16 = 1 / 2*4 = 1/8
anonymous
  • anonymous
ok but its two parts to the problem that's the answer to both?
anonymous
  • anonymous
@Preetha
anonymous
  • anonymous
@misssunshinexxoxo
anonymous
  • anonymous
@peachpi
anonymous
  • anonymous
@xapproachesinfinity
anonymous
  • anonymous
@dan815
anonymous
  • anonymous
@Compassionate
anonymous
  • anonymous
@uri
anonymous
  • anonymous
@surjithayer
anonymous
  • anonymous
@LynFran
LynFran
  • LynFran
ok what exactly u dont understand here?
anonymous
  • anonymous
i dont understand the problem or the question this whole thing confuses me.
LynFran
  • LynFran
ok lets go through this step by step from the beginning ok
anonymous
  • anonymous
ok
LynFran
  • LynFran
first of all the slope of a tangent line is the gradient of that line ok
anonymous
  • anonymous
what is a gradient
LynFran
  • LynFran
now to find the slope of the tangent line is y/x...|dw:1435279410553:dw|
LynFran
  • LynFran
The Gradient (also called Slope) of a straight line shows how steep a straight line is.
anonymous
  • anonymous
the question has two parts is 4 the answer for both
LynFran
  • LynFran
|dw:1435279812417:dw|
LynFran
  • LynFran
4 is just the slope of the tangent line we have to find the slope intercept equation at the given point ok to complete this question ok
anonymous
  • anonymous
ok whats the next step
LynFran
  • LynFran
|dw:1435280154077:dw|
LynFran
  • LynFran
what equation u get??
anonymous
  • anonymous
y=4x+2
LynFran
  • LynFran
correct
anonymous
  • anonymous
ok so thats the second part to the first question?
LynFran
  • LynFran
we just did 2nd question and yes that was this part; .Find the slope-intercept equation of the tangent line to the graph of f at the given point.
anonymous
  • anonymous
ok so 1.4 and 2. y=4x+2
LynFran
  • LynFran
yes Find the slope of the tangent line to the graph of f at the given point. is 4
LynFran
  • LynFran
for that question
anonymous
  • anonymous
ok got it
LynFran
  • LynFran
Did u do the other question?
anonymous
  • anonymous
can you help me with that one too?
LynFran
  • LynFran
ok
LynFran
  • LynFran
|dw:1435281016248:dw|
LynFran
  • LynFran
|dw:1435281095053:dw|
LynFran
  • LynFran
now to take the derivative of it we get|dw:1435281258598:dw|
LynFran
  • LynFran
|dw:1435281332523:dw|
LynFran
  • LynFran
|dw:1435281411275:dw|
anonymous
  • anonymous
ok im with you, so whats next
LynFran
  • LynFran
well we have to plug in the value of x into the derivative we just found
LynFran
  • LynFran
(16,4) and x=16 there ok so.. plug it into the derivative and let me know what u get
LynFran
  • LynFran
|dw:1435282159975:dw|
anonymous
  • anonymous
ok so when i plug in 16 and solve to find the derivative i get 0
LynFran
  • LynFran
no u cant get 0 lets work it together
anonymous
  • anonymous
ok
LynFran
  • LynFran
|dw:1435282473866:dw|
LynFran
  • LynFran
so 1/8 is this part Find the slope of the tangent line to the graph of f at the given point.
anonymous
  • anonymous
ok
LynFran
  • LynFran
so can u find this part now .Find the slope-intercept equation of the tangent line to the graph of f at the given point. |dw:1435282844490:dw|
anonymous
  • anonymous
4-y1=1/8(16-x1)
LynFran
  • LynFran
no we leave the y and x term alone and uses the y1 and x1
LynFran
  • LynFran
|dw:1435283138396:dw|
anonymous
  • anonymous
ok y-4=1/8(x-16)
LynFran
  • LynFran
correct now the next step is to multiply out the bracket
anonymous
  • anonymous
y-4=1/8x -2
LynFran
  • LynFran
good next step is to bring the -4 over the equal sign ...remember when we bring a # over the equal sign , the sign of the # changes
anonymous
  • anonymous
y=1/8x+2
LynFran
  • LynFran
yes and u have officially completed the question ok now i would like to share this site with u it was very useful to me so https://www.symbolab.com/solver/calculus-calculator/%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft(%5Csqrt%7Bx%7D%5Cright)/?origin=button
anonymous
  • anonymous
ok thank You so much for your help.
LynFran
  • LynFran
ur welcome

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