Question

How do I find the vertex of `f(x)=x^2-4x+3`?

Answer 1

We must take our equation and put it in vertex form, `f(x)=a(x-h)^2+k` where `h,k` is our vertex.

This equation will not factor into two squares `(x-h)^2` so we must use the completing the square method.

First we'll get rid of the `3` by subtracting it giving `x^2-4x=-3`

Next we take half of the -4 the square it and add it to our equation to give us `x^2-4x+4`. Since we added `4` to the left, we must also add it to the right giving `x^2-4x+4=+4-3`.

Now we can factor the left into two perfect squares, `(x-2)^2=1.`

Finally, we subtract `1` from both sides giving `(x-2)^2-1`. Our vertex is `h,k` giving `2,-1`

Answer 2

The easiest way to determine the vertex of this parabola is using a formula.

for `f(x) = ax² + bx +c`

`h = -b/(2a)`

From the function, `f(x) = x² - 4x + 3`

a = 1 b = -4 c = 3

`h = -(-4)/(2*1)`

`h = 2`

to get k, substitute the value of h for x into the function

therefore, `k = -1`