Can someone check my answer?

For y = x2 - 4x + 3,

Determine if the parabola opens up or down.

State if the vertex will be a maximum or minimum.

Find the vertex.

Find the x-intercepts.

Describe the graph of the equation.

Show all work and use complete sentences to receive full credit.,

The equation for a parabola is y = ax^2 + bx + c In this example, y = x^ 2 - 4 x + 3 a = 1 b = - 4 c = 3 Because "a" is a positive number, the parabola will open upwards and the vertex will be a minimum (at the bottom). To find the vertex, use x = - b/2a x = - (-4) / 2 (1) x = 4 / 2 x = 2 Then solve for y using x = 2 y = x^ 2 - 4 x + 3 y = 2^2 - 4 (2) + 3 y = 4 - 8 + 3 y = - 1 Therefore the vertex (x, y) is at (2, - 1) To find the x intercept, let y = 0 and solve for x. y = x^ 2 - 4 x + 3 0 = x^ 2 - 4 x + 3 0 - (x^2 - 4x + 3) = x^2 - 4x + 3 - (x^2 - 4x + 3) - x^2 + 4x - 3 = 0 Factor left side of equation: (-x + 1) (x - 3) = 0 Therefore - x + 1 = 0 or x - 3 = 0 x = 1 or x = 3 The x intercepts are (1,0) and (3,0).