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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.,

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Kalahira
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).

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