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

Mar 22, 2016

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 $0$.

${x}^{2} - 6 x + 5 = 0$

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

$\left(x - 5\right) \left(x - 1\right) = 0$

$x = 5 \text{ "" } x = 1$

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 $x = 0$:

$y = {0}^{2} - 6 \left(0\right) + 5 = 5$

There is a $y$-intercept at $\left(0 , 5\right)$, 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

$a {x}^{2} + b x + c$

can be found through the formula

${x}_{\text{vertex}} = - \frac{b}{2 a}$

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

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

${x}_{\text{vertex}} = - \frac{- 6}{2 \cdot 1} = \frac{6}{2} = 3$

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:

$y = {3}^{2} - 6 \left(3\right) + 5 = 9 - 18 + 5 = - 4$

So we know $\left(3 , - 4\right)$ 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 $\left(0 , 5\right)$ is a point, $3$ units over from the vertex, we know $\left(6 , 5\right)$ 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]}