What is the Stationary point, x and y intercepts, point of inflextion of f(x)=(x-1)(x-2)^2 ?

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#y=(x-1)(x-2)^2#

To find the stationary points we need to take the derivative of the function and set it equal to #0#.

We can either expand the second term and multiply the two parts together before taking the derivative or take it in the present form which is the product of two functions. As such, we use the product rule:

#dy/dx=(x-1)(2(x-2))+(x-2)^2=(x-2)(2x-2+x-2)#

#dy/dx=(x-2)(3x-4)=0#

#x-2=0, :. x=2#

#3x-4=0, :. x=4/3#

We can plug each one into the function to find the #y# coordinates:

#x=2, :. y=0#

#x=4/3, :. y=(4/3-1)(4/3-2)^2=1/3*4/9=4/27#

The stationary points are:

#(2,0)# is the minimum

#(4/3, 4/27)# is the maximum

We take the second derivative of the function and set it equal to #0# to find the inflection point:

#(d^2y)/dx^2=(x-2)(3)+(1)(3x-4)=3x-6+3x-4=6x-10#

#6x-10=0#

#x=5/3, :. y=(5/3-1)(5/3-2)^2=2/3*1/9=2/27#

#(5/3,2/27)# is the inflection point.

We set #x=0# to find the #y#-intercept;

#y=(-1)(-2)^2=-4#

#(0,-4)# is the #y#-intercept.

We set #y=0# to find the #x#-intercepts.

#(x-1)(x-2)^2=0#

#x=1# and

#x=2# two times.

#(1,0)# is an #x#-intercept.

The double root of #x=2# means the function touches the #x#-axis at that point but does not cross it.

The graph of the function is:

To find the stationary points, differentiate the function and solve for where the derivative equals zero. The x-intercepts are where the function crosses the x-axis, which can be found by setting f(x) equal to zero and solving for x. The y-intercept is the point where the function crosses the y-axis, which is f(0). To find the point of inflection, determine where the second derivative changes sign. 1. Stationary points: Differentiate f(x) to find its derivative: f'(x) = 3(x - 2)^2 + (x - 1)(2)(x - 2) Set f'(x) = 0 and solve for x: 3(x - 2)^2 + 2(x - 1)(x - 2) = 0 x = 2 (stationary point) 2. X-intercepts: Set f(x) = 0 and solve for x: (x - 1)(x - 2)^2 = 0 x = 1 (double root) or x = 2 (single root) 3. Y-intercept: Evaluate f(0): f(0) = (0 - 1)(0 - 2)^2 = -4 4. Point of inflection: Differentiate f'(x) to find the second derivative: f''(x) = 6(x - 2) + 2(x - 1) + 2(x - 2) Determine where f''(x) changes sign: For x < 2, f''(x) < 0; for x > 2, f''(x) > 0 Thus, the point of inflection is at x = 2.