Posted by **jane** on Tuesday, November 25, 2008 at 8:16pm.

Definition of the derivative to calculate f'(x) if f(x) = square root 2x. Find the equation of the tangent & normal line to the graph of f at x=2.

Tangent line - y=2x-x

Normal line - y = 1/2x +3

I do not know whether these are the correct answers.

- HS Calculus -
**Reiny**, Tuesday, November 25, 2008 at 9:24pm
no.

y = √(2x) = (2x)^(1/2)

y' = (1/2)(2x)^(-1/2)(2) = 1/√(2x)

when x=2, y = 2, y' = 1/2

so the tangent line is

y = (1/2)x + b

which at (2,2) is 2 = (1/2)(2) + b

b = 1

so the tangent equation is y = (1/2)x + 1

the normal would have slope of -2

so at (2,2)

2 = -2(2) + b

b = 6

the normal has equation y = -2x + 6

- HS Calculus -
**jane**, Tuesday, November 25, 2008 at 9:50pm
Thank you . I think that I understand alittle better. It is oblivious that I do not understand this material.

- HS Calculus -
**jane**, Tuesday, November 25, 2008 at 9:53pm
I have one more that I really am lost at.

Question - find the points on the graph y=2x^3-3x^2-12x+20 at which the tangent is parallel to tthe x-axis. I do not even know where to begin. We were told we could use the shortcut for this one.!

Thanks again

- HS Calculus -
**Reiny**, Tuesday, November 25, 2008 at 10:17pm
If a tangent is parallel to the x-axis, wouldn't it have a slope of zero?

And isn't the slope represented by the first derivative?

so differentiate, set that equal to zero and solve for x

find the y's for those x's and simply say

y = (whatever the y values are)

let me know if you don't get y = 0 and y = 27

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