Sketch the graph y=-x^3, using this sketch y=-x^(1/3)

The first step in figuring out this question is to first determine what the y=x3 looks like.If you are not already fimilar with this function, you can use some basic principles to find out what it looks like.When x=0 we know y=0 so we know this graph crosses the origin at 0, using this information you can determine that there are 3 roots at 0 because it is (x-0)3 = x3.You also know it goes up quickly and down quickly by determining a few points such as:x=-2 ---> y=-8x=-1 ---> y=-1x=0 ---> y=0x=1 ---> y=1x=2 ---> y=8One you have determined what y=x3 looks like you use graph transformations knowledge to determine the rest.Say y=f(x) then what is y=-f(x)?It is just a reflection in the x axis.The second part of the question requires you to understand that 1/x3 = x1/3 so the two graphs are just each others reciprocals. A reciprocal just means it is a reflection in the line y=x.

Answered by Deloris O. Maths tutor

3161 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Let f(x) = 2x^3 + x^2 - 5x + c. Given that f(1) = 0 find the values of c.


The cubic polynomial f(x) is defined by f(x) = 2x^3 -7x^2 + 2x + 3. Given that (x-3) is a factor of f(x), express f(x) in factorised form.


if a^x= b^y = (ab)^(xy) prove that x+y =1


Find the equation of the tangent to the curve y=x^3-4x^2+2 at the point (3,-7)


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