Question

How do you graph the parabola `y=x^2-4x+3` using vertex, intercepts and additional points?

Answer 1

See below

Explanation:

Roots (intercepts)
Let's first solve this parabola to obtain the roots.

Factorise `x^2-4x+3`:

`(x-3)(x-1)`
`x-3=0, x-1=0`
`x=3, x=1`

Therefore, the roots are `x=3` and `x=1`

Vertex

The vertex can be found by using the formula `-b/(2a)`. In our case, `a=1, b=-4`

Substitute `a=1, b= -4` into the formula:
`-(-4)/((2)(1))`
`-(-2)`
`2`

Therefore, the `x` coordinate of the vertex is at `x=2`

*Note that since we already knew the two roots, we could have easily gotten the `x` coordinate of the vertex by taking the average of the two roots. In this case, it is `(3+1)/2=2`. I solved it by this formula because it is important to know as well.

Substitute `2` into the original equation:

`2^2-(4)(2)+3`
`4-8+3`
`-1`

Therefore, the `y` coordinate of the vertex is at `y=-1`

Then you can roughly sketch out the graph using the two roots and the vertex.

Here is the graph:
graph{x^2-4x+3 [-10, 10, -5, 5]}

For better accuracy, you can always make a table of values. But having 3 points is usually enough.