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How do you find the slope of the indicated point for y=x^2 -4x at x=1
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y'=2x-4 y'(1)= 2(1)-4 =-2 slope=-2
slope is dy/dx at x1 so dy/dx= 2x-4 at x1=4 slope= 2-4=-2
and how would one find the equation of the tangent of this function? and moreover the equation of the normal?
\[f(a+h)-f(a)/h\] my teacher suggests that we use this expression, why is that?
i cant say why that expression is used,but since slope of the curve at x=1 is -2 , slope of tangent would be 1/2 and it passes through (1,-3). the y coordinate -3 is the value of f at x=1 so its equation is y+3=0.5(x-1)
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well he used it as a limit, this is still a bit foggy but maybe its the tiredness, thanks for the clarification
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