Find the turning points of the curve (x^3)/3 + x^2 -8x + 5

Step one, we need to assess what the question is asking us to do, in this case, find the turning points.The turning points on the graph are where the gradient is equal to zero, so in order to find the turning points, we need to se the gradient of the graph to zero.Step 2 therefore is to find the gradient, we do this by differentiation. Remember, power down in front, one off the powerd/dx x3/3 + x2 - 8x + 5 = x2 + 2x -8Step 3 is to set the gradient we have found equal to zero and solve for xx2 + 2x -8 = 0in this case, we can factorise, what could you use if you couldn't factorise? (quadratic formula)(x - 2)(x + 4) = 0therefore, x = 2 or x = -4 have we answered the question? Not yet, we need to find the corresponding y values to our x values. Step 4, We plug our x - values back into our equationfor x = 2,y = 23/3 + 22 - 82 + 5 = -4.33for x = -4y = (-4)3/3 +(-4)2 - 8(-4) + 5 = 31.67
Finally, state the answer:turning points: (2, -4.33) and (-4, 31.67)


EM
Answered by Ellie M. Maths tutor

2932 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Find the binomial expansion of (4-8x)^(-3/2) in ascending powers of x, up to and including the term in x^3. Give each coefficient as a fraction in its simplest form. For what range of x is a binomial expansion valid?


Find the derivative of yx+5y-sin(y) = x


Integrate the function f(x) = 1/(4x-1)


(a) Use integration by parts to find ∫ x sin(3x) dx


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 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences