The parabola has a minimum, the vertex and the y-intercept, you can graph it as you can see in the image attached.
Given ,
Graph of f(x) = \(x^{2}-2x+ 3\).
According to the question,
The graph of a quadratic function is a parabola whose axis of symmetry is parallel to the y -axis. The coefficients a, b, and c in the equation \(ax^{2} + bx +c = 0\).
The vertex of the parabola (It is a parabola because the equation is quadratic equation as following.
y = \(x^{2} - 2x + 3\)
\(x = \frac{-b}{2a} \\\\x = \frac{- (-2)}{2(1)}\)
x = 1
Then , \(y =1^{2} - 2(1) + 3 = 2\)
The point of the vertex at x intercept x = 1 , y = 2 (1,2) .
The sign of a is positive, therefore, the parabola has a minimum.
Substitute x = 0 into the equation to find y-intercept,
\(y = 0^{2} - 2(0) +3\)
y = 3 at x = 0
These are the points on y intercept is (0,3)
The parabola has a minimum, the vertex and the y-intercept, you can graph it as you can see in the image attached.
For more information about the Parabola click the link given below.
https://brainly.com/question/21685473
