How do you graph #F(x)=x^2-6x+8#?

1 Answer
marfre
Jul 21, 2018

See answer below

Explanation:

Given: #f(x) = x^2 -6x + 8#

This function is a quadratic - a graph of a parabola.

Factor and let #f(x) = 0# to find the x-intercepts :

#x^2 -6x + 8 = (x - 2)(x - 4) = 0#

#x - 2 = 0 " "=> x = 2; " "x - 4 = 0 " "=> x = 4

#x#-intercepts: " "(2, 0), (4, 0)#

Find the y-intercept let #x = 0#:

#f(0) = 0^2 -6*0 + 8 = 0#

#y#-intercept: " "(0, 8)#

Find the vertex . When the equation is in #Ax^2 + Bx + C = 0#,

the vertex is #(-B/(2A), f(-B/(2A)))#

#-B/(2A) = 6/2 = 3#

#f(3) = 3^2 -6*3 + 8 = -1#

vertex: #(3, -1)#

Plot a couple of other points using point-plotting. Since #x# is the independent variable, you can select any #x# and calculate the corresponding #y#:

#ul(" "x" "|" "y" ")#
#" "1" "|" "3" "#
#" "5" "|" "3" "#
#" "6" "|" "8" "#

graph{x^2 -6x + 8 [-5, 10, -2, 10]}

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