How do you use the important points to sketch the graph of #y = x^2 − 6x + 1#?

1 Answer
Feb 8, 2018

#color(blue)("Y-int": (0,1))#
#color(red)("Vertex": (3,8))#
#color(green)("X-Intercept/Zero":(0.172,0) and (5.828,0)#

Explanation:

The most important points should be the #color(red)"vertex"#, the #color(blue)("y-int")#, and the #color(green)"zeroes"# (if there are any).

The x-coordinate of the #color(red)(vertex)# of any quadratic equation
#y=ax^2+bx+c# is:

#(-b)/(2a)#

Plug in the x-coordinate back into the equation to find y. Do it on this equation:

#b=-6#

#a=1#

#c=1#

#(-(-6))/(2a)=3#

Now you have the #color(red)(vertex)# as #(3,y)#

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

The #color(red)(vertex)# is #(3, -8)#

The #color(blue)("y-int")# of the quadratic equation
#y=ax^2+bx+color(orange)(c)# is simply #color(orange)(c)#.

The #color(blue)("y-int")# of this equation is #(0,1)#.

To find the #color(green)"zeroes"#, plug into the quadratic formula, which is given by:

#(-b+-sqrt(b^2-4ac))/(2a)#

Plug in:

#(6+-sqrt((-6)^2-4*1*1))/(2*1)#

Simplify:

#(6+-4sqrt(2))/2#

#3+-2sqrt(2)#

The #color(green)"zeroes"# are: #(5.828,0) and (0.172,0)#