A curve is given by the equation y=x^3-11x^2+28x; find the coordinates of the points where the curve touches the x-axis.

Since the equation has a cubic term, we can expect the curve to have certain properties: it should have three points where the graph touches the x-axis, two turning points, of which one should be a minimum and one should be a maximumTo start we will find the point where the curve touches the x-axis and to do that we have to simplify the equation of the curve by factorising it, since every term has an x value, we can factorise the whole equation by x, giving us x(x^2-11x+28). Now we have a quadratic equation which we can factorise by sight or by using the quadratic equation, to give us y=(x)(x-4)(x-7), this tells us that when the curve touches the x-axis (when y=0) x = 0, 4, 7, therefore the coordinates are (0,0), (4,0) and (7,0)

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Answered by Lui H. Maths tutor

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