How do you find the vertex and axis of symmetry, and then graph the parabola given by: #y = x^2 - 5x + 3#?

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Answer 1 · 2018-05-14T02:27:15

#"vertex" = (5/2, -13/4); " axis of symmetry": x = 5/2#

Given: #y = x^2 -5x +3#

Standard form of a parabola: #y = a(x-k)^2 +k#

where #"vertex" = (h, k); "axis of symmetry": x = h"

and #a = " constant"#

The vertex can be found easily: #"vertex" = (-B/(2A), f(-B/(2A)))#

where the equation is in general form: #Ax^2 + Bx +C = 0#

#-B/(2A) = 5/2#

#f(-B/(2A)) = f(5/2) = (5/2)^2 -5(5/2) + 3#

# = 25/4-25/2 + 3 = 25/4 - 50/4 + 12/4 = -13/4#

#"vertex" = (5/2, -13/4)#

# " axis of symmetry": x = 5/2#

You can also do competing of the square to put the equation in standard form.

#y = (x^2 -5x) +3#

#y = (x - 5/2)^2 + 3 - (5/2)^2#

#y = (x - 5/2)^2 + 12/4 - 25/4#

#y = (x - 5/2)^2 - 13/4#

graph{x^2-5x+3 [-10, 10, -5, 5]}