How do I find the tangent line of y=f(x) at x=2?
To find the tangent line at x = 2, you would need to find the derivative of f and substitute x = 2. That gives you the slope of the tangent line. To write the equation of the line, you would need a single point that the curve f passes through.
If f(2)=3 and f'(2)=5, find an equation of (a) the tangent line , and (b) the normal line to the graph of y=f(x) at the point where x=2. that is the exact question
I will start to help you with part(a). Well you know that f ' (2) = 5 and (2,3) is a point on the curve of f, so the equation of the tangent line is y = 3 = 5(x - 2).
In part(b), you just want the equation of the normal line. The normal line is the line perpendicular to the tanglent line...so that's the same idea as part (a), but the slope is the negative reciprocal of 5.
So now you have it all.
How did you know that was the equation of the the tangent line?
From algebra, the equation of the line is y - y1 = m(x - x1). so since m = 5 (you were told that f ' (2) = 5), and the curve passes through (2,3)..so x = 2 and y = 3. So substitute into the equation of the line.
ohhh. thank you very much
welcome.
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