Answer:
part A: x= 1 and x= 3/2
part B: minimum at (5/4, f(5/4))
Explanation:
part A: factoring
0 = 2x^2 - 5x + 3
= (2x^2 - 2x)(-3x + 3)
= 2x(x - 1) -3(x-1)
= (2x - 3)(x - 1)
x= 1 and x= 3/2
part B: candidates test
find the derivative of the function.
f(x) = 2x^2 - 5x + 3
f'(x) = 4x -5
solve for critical numbers.
0 = 4x - 5
x = 5/4
candidates test by plugging in numbers into the derivative to see if that area is negative or pos.
f has a minimum at x = 5/4 because f' goes from neg to pos there. to find the coordinate for x = 5/4 just plug it into the original function to find f(5/4).
Part C: to draw the graph, I'd just plot the vertex coordinate that you solved for in part B, then I would plot the x intercepts that you solved for in part a, from there just draw a line throu the 3 points. (it should look like a U shape)