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.

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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.

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find the derivative f(x) = 4x + 2 can you find f'(x)?
-1/2+x/4

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no the derivative of that is 4
yes thats right so the slope is - at every point on the graph as its a straight line
* slope is 4
the second function is:- f(x) = x^(1/2) can you find the derivative of this one?
x^-1/2/2
|dw:1435264790652:dw|
so to find the slope at (16,4) plug x = 16 into the the above value for f'(x)
then use the general form y-y1 = m(x-x1) and plug in slope m and the point (x1,y1) will be (16,4)
for the first step is it 2
1 / 2*x(1/2) = 1 / 2* (16)^1/2 = 1/ 2*4 = 1/8
so m = 1/8
im lost. I don't understand this at all.
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
ok but its two parts to the problem that's the answer to both?
ok what exactly u dont understand here?
i dont understand the problem or the question this whole thing confuses me.
ok lets go through this step by step from the beginning ok
ok
first of all the slope of a tangent line is the gradient of that line ok
what is a gradient
now to find the slope of the tangent line is y/x...|dw:1435279410553:dw|
The Gradient (also called Slope) of a straight line shows how steep a straight line is.
the question has two parts is 4 the answer for both
|dw:1435279812417:dw|
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
ok whats the next step
|dw:1435280154077:dw|
what equation u get??
y=4x+2
correct
ok so thats the second part to the first question?
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.
ok so 1.4 and 2. y=4x+2
yes Find the slope of the tangent line to the graph of f at the given point. is 4
for that question
ok got it
Did u do the other question?
can you help me with that one too?
ok
|dw:1435281016248:dw|
|dw:1435281095053:dw|
now to take the derivative of it we get|dw:1435281258598:dw|
|dw:1435281332523:dw|
|dw:1435281411275:dw|
ok im with you, so whats next
well we have to plug in the value of x into the derivative we just found
(16,4) and x=16 there ok so.. plug it into the derivative and let me know what u get
|dw:1435282159975:dw|
ok so when i plug in 16 and solve to find the derivative i get 0
no u cant get 0 lets work it together
ok
|dw:1435282473866:dw|
so 1/8 is this part Find the slope of the tangent line to the graph of f at the given point.
ok
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|
4-y1=1/8(16-x1)
no we leave the y and x term alone and uses the y1 and x1
|dw:1435283138396:dw|
ok y-4=1/8(x-16)
correct now the next step is to multiply out the bracket
y-4=1/8x -2
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
y=1/8x+2
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
ok thank You so much for your help.
ur welcome

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