How do you graph #y=x^2- 6x+2#?

Question

5206 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-22T02:16:48

#"y-int" = 2#
#"x-int" = 3+-sqrt7#
vertex = #(3,-7)#

#y=x^2- 6x+2#

your function is in the form: #ax^2+bx +c#

#a=1#
#b=-6#
#c=2#

From this we can derive:

y-intercept = c

y-int = 2

x-intercept(s) if they exist are the solutions or roots:

#y=x^2- 6x+2#

Roots (found with the quadratic formula): #3+-sqrt7#

axis of symmetry is: #aos=(-b)/(2a) = (-(-6))/(2*1)=3#

vertex is: #(aos, "*"f(aos))#
*this means you put the aos value back into the function as x and solve for y.

vertex is: #(3, 3^2- 6*3+2)#

vertex is: #(3,-7)#

Finally here is your graph:

graph{x^2- 6x+2 [-9.84, 10.16, -7.48, 2.52]}