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

Dec 1, 2014

We must take our equation and put it in vertex form, $f \left(x\right) = a {\left(x - h\right)}^{2} + k$ where $h , k$ is our vertex.

This equation will not factor into two squares ${\left(x - h\right)}^{2}$ so we must use the completing the square method.

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

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

Now we can factor the left into two perfect squares, ${\left(x - 2\right)}^{2} = 1.$

Finally, we subtract $1$ from both sides giving ${\left(x - 2\right)}^{2} - 1$. Our vertex is $h , k$ giving $2 , - 1$

Dec 6, 2014

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

for  f(x) = ax² + bx +c

$h = - \frac{b}{2 a}$

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

a = 1 b = -4 c = 3

$h = - \frac{- 4}{2 \cdot 1}$

$h = 2$

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

therefore, $k = - 1$