# How do you find the vertex and axis of symmetry, and then graph the parabola given by: y = x^2 - 4x + 1?

May 20, 2018

$a o s = 2$
$v e r t e x = \left(2 , - 3\right)$
$\text{y-int} = 1$
$\text{x-int} = 2 - \sqrt{3} \mathmr{and} 2 + \sqrt{3}$

#### Explanation:

Standard form is:

$f \left(x\right) = a {x}^{2} + b x + c$

$a = 1$
$b = - 4$
$c = 1$

The formula for the axis of symmetry is:

$a o s = \frac{- b}{2 a}$

$a o s = \frac{- \left(- 4\right)}{2 \cdot 1}$

$a o s = 2$

Formula for the vertex point $\left(x , y\right)$ in:

$v e r t e x = \left(a o s , f \left(a o s\right)\right)$

what is f(aos)? that means you put the value you found for the aos back into the function as x and solve for y:

$y = {x}^{2} - 4 x + 1$

$y = {2}^{2} - 4 \left(2\right) + 1$

$y = - 3$

so the vertex is: $\left(2 , - 3\right)$

Finally to help you graph the y-intercept is $c$ so:

y-int = 1

also the x-intercepts are the roots if you factor the polynomial, you would need to use the quadratic formula to factor this one and the roots will be:

$x = 2 + \sqrt{3}$
and
$x = 2 - \sqrt{3}$