The points (0, -5) and (5, 0) lie on a curve y=x^2 + ax + b. Find the stationary points on the curve.

Firstly, let us try to calculate the values for a and b to obtain an equation for the curve. To do this we will substitute the coordinates we are given into y=x2 + ax + b:Substituting in (0, -5) -> -5 = (0)2 + a(0) + b -> b = -5Substituting in (5, 0) -> 0 = (5)2 + a(5) + b = 20 + 5a -> 5a = -20 -> a = -4Hence, we have that the curve is given by the equation y = x2 - 4x - 5As curve is described by a quadratic equation it must be a parabola. A feature of all parabolas are that they are symmetrical about their minimum, which is also their only turning (stationary) point. This means that the x coordinate of the stationary point on the curve will be halfway between the two roots of the curve. We know one of the roots is (5, 0), by factorising, y=x2 - 4x - 5 = (x-5)(x+1), we can see that the other root occurs at x=-1. Therefore, the turning point occurs at x=(5 - 1)/2 = 2.Finally, we can obtain the y coordinate for the turning point by substituting into the equation for the curve at the point x=2: y = (2)2 -4(2) - 5 = 4 - 8 - 5 = -9. So, the stationary point on the curve occurs at (2, -9).

Answered by Adam R. Maths tutor

4223 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

solve this equation x^2 -2x=-1


Solve 4(x+3)=2x+8


Solve the simultaneous equations: 4x + y = 25, x - 3y = 16


Solve 3(2x + 5) = 4 – x


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