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)

Answered by Lui H. Maths tutor

2506 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

What is compound interest?


Simplify: 5a + 2 – a + 9


Solve the equation 18x^2-3x=6


What is the size of the exterior and interior angle of a regular 13 sided polygon?


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2024

Terms & Conditions|Privacy Policy
Cookie Preferences