Graph quadratic fuctions in standard form

Question

Graph the fuction by hand. Find the vertex round approximate coordinates to the second decimal place.

y=x^2-4x+3

Answer 1

f(x) = y = x^2 -4x + 3

y intercept is 3, the constant term, the point (0,3)

complete the square

= x^2 -4x + 4 = 4-3 = 1

(x-2)^2 = 1

take the square root of both sides

x-2 = 2 + or - 1 = 3 or 1 = x intercepts: (1,0) and (3,0)

vertex is (2, -1)= the minimum point of an upward opening parabola

axis of symmetry is x=2

so another point on the graph is (4,3) symmetric to the y intercept (0,3) about the x=2 line

another way to find the vertex is take the derivative and set it =0

f'(x) = 2x-4 =0

x = 2 = x coordinate of the vertex

plug 2 into the original quadratic

2^2 -4(2) + 3 = -1 = y coordinate of the vertex