if f(2)=3 and f'(2)=5 find and equation of the tangent line and the normal line to the graph of y=f(x) I know satellite73 answered this, but I am curious what the normal line's equation is and how to find it
Normal line is the line perpendicular to the tangent line and goes through the point at (2,3).
Or whatever point of tangency we are talking about (2,3) being the one here.
Okay, so the inverse of the tangent line: y=5x-7
What's the easiest way to find the inverse?
You want to find the inverse now?
Yeah because I already have found the tangent line.
Since lines are 1 to 1, all you need to do is solve for x. Replace x with f^(-1)(x) and y with x. I think you are confused. Why are we finding the inverse of the line?
Sorry, I thought that was just another name for the normal line... >.<
Nope :p
So after we find the slope of the tangent line. Let's call that slope m. Then the slope of the normal line with be -1/m.
Lines have form y=slope*x+b
The slope of the normal line is -1/m So we have y=-1/m*x+b
So it's just -1/5x-7?
Did you put your point into find the y-intercept?
o.o
y=-1/m*x+b The point this lines goes through or should go through is (2,3) put that point in 3=-1/m*2+b solve for b. 3=-2/m+b 3+2/m=b So you have y=-1/m*x+(3+2/m)
So, y=-1/5x+17/5 ?
yep
Cool ^w^ Thank you!
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