Question

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

Answer 1

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]}