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Find all points of the graph of 
y=x^(3)-2x^(2)+1 where the gradient equals the 
y-coordinate.
[5 marks]

88. Find all points of the graph of y=x32x2+1 y=x^{3}-2 x^{2}+1 where the gradient equals the y y -coordinate.\newline[55 marks]

Full solution

Q. 88. Find all points of the graph of y=x32x2+1 y=x^{3}-2 x^{2}+1 where the gradient equals the y y -coordinate.\newline[55 marks]
  1. Find Derivative: First, find the derivative of yy with respect to xx to get the gradient of the graph.dydx=3x24x\frac{dy}{dx} = 3x^2 - 4x
  2. Set Gradient Equal: Set the gradient equal to the y-coordinate. 3x24x=x32x2+13x^2 - 4x = x^3 - 2x^2 + 1
  3. Rearrange Equation: Rearrange the equation to bring all terms to one side and set it equal to zero.\newlinex32x23x2+4x1=0x^3 - 2x^2 - 3x^2 + 4x - 1 = 0\newlinex35x2+4x1=0x^3 - 5x^2 + 4x - 1 = 0
  4. Solve for xx: Now, we need to solve for xx. This is a cubic equation, and it might be tough to solve by hand. We can try factoring or use the Rational Root Theorem, or synthetic division, or graphing calculator to find the roots.
  5. Factor or Use Methods: Let's try to factor by grouping or see if there's an obvious factor. x35x2+4x1x^3 - 5x^2 + 4x - 1 doesn't factor nicely, and there's no obvious root. We might need to use a graphing calculator or numerical methods to find the roots.
  6. Plug in Values: Assuming we found the roots of the equation x35x2+4x1=0x^3 - 5x^2 + 4x - 1 = 0 to be x1x_1, x2x_2, and x3x_3, we would then plug these values back into the original equation y=x32x2+1y = x^3 - 2x^2 + 1 to find the corresponding yy-coordinates.

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