At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
@dan815 could you help me?
the derivative formula that I got is f'(x)= -6x+1
but none of the answers match my answer
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
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
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
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) = ?
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?
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
I got 10
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=
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?
I already found that C actually has the value, but I don't know how to solve A nor B
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)
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
okay, I see. I got 7 as the slope of the function
if we keep making h smaller and smaller then the point is close and closer to the actual tangent slope
which is 7 right?
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.