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Find the equation(s) of the tangent line(s) for the graph f(x)=2x^2-3x-5 that contain the point (1,-8). I have f'(x)=4x-3... now what do I do? Btw, (1,-8) is not a point on the graph.
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in f'(x)=4x-3 find f'(1) ===> that weill give the slope of the tangent ===> write the line formula y = mx+b ==> subititue m(slope) and (1,-8) to get b , that will be the the equation of the tangent line. now , regarding what you have mentioned about the point (1,-8) , then there must be amistake in the given data, but what is importany here is the idea.
The graph of the equation is a parabola, and the point (1,-8) is below the graph, so there should be two possible lines tangent to the graph that goes through the point (1,-8).
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