How do you find the vertex and axis of symmetry, and then graph the parabola given by: #y = x^2 - 4x + 1#?

1 Answer
Bill Jorgensen
May 20, 2018

#aos = 2#
#vertex = (2, -3)#
#"y-int" = 1#
#"x-int" = 2-sqrt3 and 2+sqrt3#

Explanation:

Standard form is:

#f(x)=ax^2+bx+c#

your equations is already in standard form.

#a=1#
#b= -4#
#c =1#

The formula for the axis of symmetry is:

#aos = (-b)/(2a)#

#aos = (-(-4))/(2*1)#

#aos = 2#

Formula for the vertex point #(x,y)# in:

#vertex = (aos, f(aos))#

what is f(aos)? that means you put the value you found for the aos back into the function as x and solve for y:

#y= x^2-4x+1#

#y = 2^2-4(2) +1#

#y = -3#

so the vertex is: #(2, -3)#

Finally to help you graph the y-intercept is #c# so:

y-int = 1

also the x-intercepts are the roots if you factor the polynomial, you would need to use the quadratic formula to factor this one and the roots will be:

#x = 2+sqrt3#
and
#x = 2-sqrt3#

Related questions
See all questions in Quadratic Functions and Their Graphs
Impact of this question
1368 views around the world
You can reuse this answer
Creative Commons License