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

##### 1 Answer

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

**Determine the function's zeros:**

We can find where the function crosses the

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

Factor this by looking for two numbers that add up to

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

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

We know the graph of the parabola will cross the

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

There is a

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

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

The

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

So we know

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

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]}