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Once you find f'(x), plug in a to get the slope of the line, m. Then you want to get it in the form y=mx+b, by using the fact it passes through the point (a, f(a))
Do you know how to take the derivative? If so, then first take the derivative of the function, then evaluate it at the given a -- this is "m", the slope of the tangent line. Now, evaluate the function at the given a. This we'll call "y1" The tangent line is: y-y1 = m (x-a)
Can please someone tell me what is a slope just so I get the bigger picture of whats going on?
Slope is the steepness of the line, or rate of change. Remember back in algebra, equations y=mx+b have a slope of m.
@BTaylor what do you mean by "Now, evaluate the function at the given a. " ?
so, for number 9, f(x) = 2x^2. The given "a" is 1. So evaluating the function at a means you plug in a for x, and get 2(1)^2, which is 2.
That means plug in x=1 (since a=1) into f = 2x^2
@agent0smith is right. To visualize this tangent line, click this link: http://www.wolframalpha.com/input/?t=crmtb01&f=ob&i=tangent%20line%20to%202x%5E2%20at%20x%3D1
I got y1 = 0
y1 = -2
y1 = 2 :D **
the last one is right. y1 = 2.
so that's the answer?? y1 = 2 and I am done? Nothing else needed?
No... did you find the slope?
yes. because the derivative is 4x, evaluated at x=1 gives you 4(1) =4
So then put it in point-slope form: y-y1 = m (x-x1) where x1 = 1, y1= 2
I did all that and I got y1 = 2 in the end, I am done ?
Wait, no, where is your x going? y-y1 = m (x-a) y - 2 = 4(x-1)
x-1 = 0
x = 1
x is x.
a is 1.
then why y is not y too and had to be replaced
y is y. y-2 = 4 (x-1)
You're using point slope form to write the equation of a line: http://www.purplemath.com/modules/strtlneq2.htm
Nice! I see now! Thank you both! I wish I could give more medals!
Your line has a slope of m=4, and passes through the point (1, 2) then you just plug it into the point slope formula.
oh i see
And again, this is what you have found: http://www.wolframalpha.com/input/?t=crmtb01&f=ob&i=tangent%20line%20to%202x%5E2%20at%20x%3D1