Find the equation of the line tangent to y = –x2 + 3x + 8 at x = 2.

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Find the equation of the line tangent to y = –x2 + 3x + 8 at x = 2.

Mathematics
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@dan815 could you help me?
the derivative formula that I got is f'(x)= -6x+1
but none of the answers match my answer

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Other answers:

Find the equation of the line tangent to y = –x2 + 3x + 8 at x = 2. y = –2x + 2 y = –x + 12 y = –x + 2 y = –2x – 12
|dw:1436827395674:dw|
|dw:1436827520119:dw|
y = –x + 12
is it 26
could you explain me how you got that answer
did you get the derivative function right
yes, because when I plug the value of 2 into my formula I get -1, which you presented me that it was correct, but I don't know how you got that formula
f'(x)= -6x+1 you said you got this, something wrong there
you have to differentiate term by term
never mind I got -2x+3
okay
now you have a slope, then you find a point that belongs on this tangent line
but that doesn't have anything in common to the other answers
how?
|dw:1436828112090:dw|
we know the x coordinate of the point,
now we find the y coordinate of the point and its part of the parabola so
y(2) = ?
|dw:1436828189830:dw|
do we plug it into the original formula or the derivate one?
we have the slope there, now we need to find the y coordinate then you will have a point and a slope and you figure out the equation of the line
the derivative formula only tells you the slope at every x value
we want to know the actual y point
so which formula could I use to find the y value?
think about it
the tangent line intersects one point on the parabola, at x=2 and y is something, we can always find the y value for every x value on a parabola how?
maximum
wait no, OMG I feel so ignorant.
y = –x^2 + 3x + 8 this equation of parabola
it gives you a y value for every x value
we know there is an intersection at x=2 so we have to see the y value there
y = –x^2 + 3x + 8 y=-2^2+3*2+8 = 10
x=2,y=10 (2,10)|dw:1436828495281:dw|
I got 10
yes,
now u have a point at the slope of the tangent line through that point
can you solve the equation of the line from here
y=mx+b m=-1 lets plug in our point to see what b or the y intercept must be 10=-1*2+b b=
9
10=-1*2+b 10=-2+b b=10+2=12
Oh i see I thought it was a 1 to the second power
this is the equation of a line y=mx+b
yeah, OMG I got it. Thank you so much. Could you help me with one more and thats it?
yeah sure
I already found that C actually has the value, but I don't know how to solve A nor B
|dw:1436829534343:dw|
this is the definition of a derivative
it is the equation to find the slope of a tangent line at x
it arises from taking a secant line where the points are getting closer and closer, lim h--->0
for example take a look at any function f(x)
|dw:1436829711928:dw|
suppose you want to find the derivative at that point x for this function, you can begin to approximate by taking points close to that value because that is a close linear approximation
|dw:1436829768720:dw|
okay, I see. I got 7 as the slope of the function
at x=2
|dw:1436829791039:dw|
if we keep making h smaller and smaller then the point is close and closer to the actual tangent slope
which is 7 right?
|dw:1436829918175:dw|
well my work is done as long you understand how a derivative comes
its from this very simple concept, the slope of the tangent line at some point
okay than you, I understood the secant and tangent line more.

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