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
Thank you . I think that I understand alittle better. It is oblivious that I do not understand this material.
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.!
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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