A curve is defined by the equation y = (x + 3)(x – 4). Find the coordinates of the turning point of the curve.

The turning point of a curve is the point at which it will turn, therefore, either the maximum or minimum point. Firstly, you need to expand out the brackets so the equation looks like a standard curve equation. When expanded, the equation will be y = x2 – x – 12. The turning point will be when dy/dx is equal to 0.Therefore when the curve equation is differentiated you get 2x - 1 . Set this to 0.When solved, x will equal 0.5 . This is the x coordinate of the turning point. You need to find the y coordinate as well. To do this, fill 0.5 back into the original curve equation where an x is. The equation will now read, y = 0.25 - 0.5 - 12 . Y will equal -12.25 . Therefore the turning point will be at (0.5, -12.25) . You have now finished the question.

CM

Related Further Mathematics GCSE answers

All answers ▸

x^3 + 2x^2 - 9x - 18 = (x^2 - a^2)(x + b) where a,b are integers. Work out the three linear factors of x^3 + 2x^2 - 9x - 18. (Note: x^3 indicates x cubed and x^2 indicates x squared).


A curve is mapped by the equation y = 3x^3 + ax^2 + bx, where a is a constant. The value of dy/dx at x = 2 is double that of dy/dx at x = 1. A turning point occurs when x = -1. Find the values of a and b.


3x^3 -2x^2-147x+98=(ax-c)(bx+d)(bx-d). Find a, b, c, d if a, b, c, d are positive integers


Find the stationary points of y=x^3 + 3x^2 - 9x - 4