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)

LH
Answered by Lui H. Maths tutor

3265 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Megan buys a car for £7,700 plus 20% VAT. She pays a deposit for the car and then pays the rest in 12 equal instalments of £642.50. How much did Megan pay as a deposit?


How do I solve x-6=15?


simplify 4p^3 x 3p^4


x : y = 7 : 4, x + y =88. Work out the value of x - y.


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning