How do I graph #y=x^2-6x+5#?

If we want to take a more algebraic/analytical approach:

Determine the function's zeros:

We can find where the function crosses the #x#-axis by setting the function equal to #0#.

#x^2-6x+5=0#

Factor this by looking for two numbers that add up to #-6# and multiply to #5#. In this case, these are #-5# and #-1#.

#(x-5)(x-1)=0#

#x=5" "" "x=1#

We know the graph of the parabola will cross the #x#-axis at these two points. We can mark them on a graph:

graph{((x-5)^2+y^2-1/30)((x-1)^2+y^2-1/30)=0 [-7.33, 15.18, -6.24, 7.01]}

Determine the function's y-intercept:

The #y#-intercept will occur when #x=0#:

#y=0^2-6(0)+5=5#

There is a #y#-intercept at #(0,5)#, which we can mark on our graph.

graph{((x-5)^2+y^2-1/30)((x-1)^2+y^2-1/30)(x^2+(y-5)^2-1/30)=0 [-7.33, 15.18, -6.24, 7.01]}

Determining the vertex:

The #x#-coordinate of the vertex of a parabola in the form

#ax^2+bx+c#

can be found through the formula

#x_"vertex"=-b/(2a)#

(This is also on the vertical line where the parabola's axis of symmetry lies.)

For #x^2-6x+5#, we see that #a=1# and #b=-6#, so

#x_"vertex"=-(-6)/(2*1)=6/2=3#

The #y#-coordinate of the vertex can be found through plugging in #x=3#, which is the #x#-coordinate of the vertex, into the parabola's equation:

#y=3^2-6(3)+5=9-18+5=-4#

So we know #(3,-4)# is the vertex of the parabola.

graph{((x-3)^2+(y+4)^2-1/20)((x-5)^2+y^2-1/30)((x-1)^2+y^2-1/30)(x^2+(y-5)^2-1/30)=0 [-7.33, 15.18, -6.24, 7.01]}

Using the symmetric properties of a parabola, since we know #(0,5)# is a point, #3# units over from the vertex, we know #(6,5)# will also be on the parabola since it is #3# units over in the other direction:

graph{((x-6)^2+(y-5)^2-1/30)((x-3)^2+(y+4)^2-1/20)((x-5)^2+y^2-1/30)((x-1)^2+y^2-1/30)(x^2+(y-5)^2-1/30)=0 [-7.33, 15.18, -6.24, 7.01]}

Using these points, we can draw the parabola quite well:

graph{x^2-6x+5 [-8.36, 14.15, -5.5, 7.75]}