The slope of the tangent to the curve y=x^2 - X at the point, where the line y=2 cuts the curve in the 1st quadrant, is
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The point at which line y=2 cut the curve can be found out by
2=x^2-x
x^2-x-2=0
on solving we will get,
x=-1,2
since, the point is in first quadrant, So, x=2
Thus, point is (2,2).
Slope of the curve can be found out by differentiating the the equation of curve with respect to the x.
dy/dx= 2x-1
Now, put the value of x
dy/dx=2(2)-1=4-1=3
Therefore, the slope is 3.
2=x^2-x
x^2-x-2=0
on solving we will get,
x=-1,2
since, the point is in first quadrant, So, x=2
Thus, point is (2,2).
Slope of the curve can be found out by differentiating the the equation of curve with respect to the x.
dy/dx= 2x-1
Now, put the value of x
dy/dx=2(2)-1=4-1=3
Therefore, the slope is 3.
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