Find the coordinates of the stationary point on the curve y=2x^2+3x+4=0

The stationary point on the curve is the point on the curve where the gradient is 0. In other words the tangent to the curve at that point is horizontal. The gradient of the curve can be expressed as dy/dx which is the first differential of the curve. dy/dx in effect gives us a formula to calculate the gradient of the curve at any x value. 2x^2+3x+4 differentiates to 4x+3. Since we know at the stationary point the gradient is 0 4x+3 must be equal to 0. Rearranging for x gives us x=-3/4 so the gradient of the curve is 0 at -3/4 and there is only 1 stationary point. To find the y coordinate substitute x=-3/4 into the original equation. y=2(-3/4)^2+3(-3/4)+4=23/8. This curve has 1 stationary point with coordinates (-3/4,23/8)

Answered by Qais Z. Maths tutor

4612 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Using the addition formula for sin(x+y), find sin(3x) in terms of sin(x) and hence show that sin(10) is a root of the equation 8x^3 - 6x + 1


Find the integral of sin^2(X)


Differentiate f(x) with respect to x. Find the stationary value and state if it is a maxima, minima or point of inflection f(x) = 6x^3 + 2x^2 + 1


Explain the chain rule of differentiation


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