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Calculus/point of tangency


my teacher gave me these questions and I don't know how to solve these :)
1.) at what point on the curve y=x^3 is the tangent parallel to the line with equation y=9x
2.) at what point on the given curve y=x^3 is the tangent line parallel to the line 3x-y=3?
3.)  a. find the slope of the tangent line to the f(x) at the point (1,2)?
     b. determine the equation of the tangent line
thanks its a long quiz

The slope of a line is given by the derivative of that line,
1) If we have y = x^3, y' = 3x^2.
If we have a line y=9x, the derivative of that curve is y' = 9.
The equation to solve is 3x^2 = 9, so solve for x.
Remember that x will have a positive and negative value when taking a squareroot.

2) The slope of y=x^3 is y' = 3x^2.
The slope of the line y = 3x-3 is 3.
The equation to solve for is then 3x^2 = 3.
Again, note that x gives a plus and a minus answer when taking a squareroot.

3) Again, the slope of a tangent line at (1,2) can be found if (1,2) is on the curve.
It is f'(1).  The equation can be found by taking the point-intercept form of the line.
That is, f(x) - 2 = f'(2)(x-1).  


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