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Find all points (x,y) on the graph of f(x)=2x^2-3x with tangent lines parallel to the line y=9x+4
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we have to compute the first derivative of your function, first
do you know how to compute such first derivative?
yes i do
it would be 4x-3
ok! now using the geometrical meaning of the first derivative, we have to solve this equation: \(4x-3=9\) being \(9\) the slope of the line
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please solve that equation with respect to \(x\) what do you get?
x=3
correct!
we got only one x-coordinate, so the y-coordinate is: \(y=2\cdot 3^2-3\cdot 3=...\)
y=9
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correct! So the requested point is \((3,9)\)
It says all points is (3,9) the only one?
yes! It is the only point, since the first derivative is a polynomial of degree 1
Ok thank you very much
:)
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