At what point on the graph y=1/2x^2 is the tangent line parallel to the line?
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OpenStudy (anonymous):
what line is the question referring to?
OpenStudy (anonymous):
sorry, i missed that. It's 2x-4y=3
OpenStudy (anonymous):
ok...
so all we need is the slope of the line 2x - 4y = 3 so we can set the slope of the tangent line equal to that. did you find the derivative of y = 1/2x^2 ?
OpenStudy (anonymous):
x?
OpenStudy (anonymous):
is that the derivative?
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OpenStudy (anonymous):
Yes it is
OpenStudy (anonymous):
ok....
since the derivative gives the slope of the tangent line at any point, then all we need to do now is set that to the slope of the line 2x - 4y = 3.
what's the slope of 2x - 4y = 3 ????
OpenStudy (anonymous):
y=1/2x - 3/4
OpenStudy (anonymous):
just the slope.... what is it?
OpenStudy (anonymous):
1/2
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OpenStudy (anonymous):
ok... good... lemme summarize....
OpenStudy (anonymous):
dy/dx = x
slope = 1/2
so if we set dy/dx = slope, we have:
x = 1/2
this is the x-coordinate of the point on the graph where the tangent line is parallel to
2x - 4y = 3. do you know how to get the y-coordinate?
OpenStudy (anonymous):
Is that where we look for the x value?
OpenStudy (anonymous):
no... we already have the x value. we only need the y value (coordinate) of the point on the graph.
OpenStudy (anonymous):
hint: how do you plot points on the graph of y = 1/2x^2 ? how do you calculate the y-value when x=1/2 ?
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