please work out the turning point of the curve and the max and min points of f(x)=x2-3x+2?

f(x) = x^2-3x+2
f'(x) = 2x-3 =0
2x-3=0
x=3/2 (turning point)

f''(x) = 2 > 0 so f hasa global minimum at x=3/2
The global minimum is f(3/2) = (3/2)^2 -3(3/2) + 2
= 9/4 -9/2 + 2
= (9-18+8)/4 = -1/4

There is no maximum

f(x) = x² - 3x + 2

f'(x) = 2x - 3
f"(x) = 2

When x = 3/2 :
f'(x) = 0
f"(x) = 2 > 0
Hence, the minimum point at x = 3/2.

Minimum f(x) = (3/2)² - 3(3/2) + 2 = -1/4

Hence, minimum point at (3/2, -1/4)

f(x) = x^2 - 3x + 2
turning point when f'(x) = 0

f'(x) = 2x - 3
f'(x) = 0
2x - 3 = 0
x = 3/2

f(3/2) = (3/2)^2 - 3(3/2) + 2
f(3/2) = -1/4
turning point at(3/2, -1/4)

A parabola, a > 0 : minimum at turning point.

f'(x) = 2x-3 = 0
Minimum at (3/2, -1/4)
Maximum value ∞

f ` (x) = 2x - 3
f " (x) = 2
Thus a Minimum turning point.

For turning point
2x - 3 = 0
x = 3/2
f (3/2) = 9/4 - 9/2 + 2
f (3/2) = 9/4 - 18/4 + 8/4 = - 1/4

Minimum Turning point (3/2 , - 1/4)
As x --> ∞ , f (x) --> ∞
As x --> - ∞ , f (x) --> ∞

Max and min occur when the first derivative is zero.
Inflection points occur when the second derivative is zero.

You have calculated the derivative as a function of x. Write the equation [that function] = 0. Solve for x. Then put that x back into the original equation to find y.

This equation is simple enough that you should know approximately what the graph looks like, and hence whether critical points are maximum or minimum. But in general: at a maximum, the second derivative is negative, and at a minimum, the second derivative is positive.