# How do you graph the function, label the vertex, axis of symmetry, and x-intercepts. y=x^2-2x+1?

Aug 4, 2018

Vertex is at $\left(1 , 0\right)$ ,axis of symmetry is $x = 1$
x intercept is $x = 1$ ,y intercept is $y = 1$, additional
points on parabola are
$\left(- 1 , 4\right) , \left(3 , 4\right)$

#### Explanation:

$y = {x}^{2} - 2 x + 1 \mathmr{and} y = {\left(x - 1\right)}^{2} + 0$

This is vertex form of equation ,y=a(x-h)^2+k ; (h,k)

being vertex , here $h = 1 , k = 0 , a = 1$

Since $a$ is positive, parabola opens upward.

Therefore vertex is at $\left(h , k\right) \mathmr{and} \left(1 , 0\right)$

Axis of symmetry is $x = h \mathmr{and} x = 1$

x-intercept is found by putting $y = 0$ in the equation

$y = {\left(x - 1\right)}^{2} \mathmr{and} {\left(x - 1\right)}^{2} = 0 \therefore x = 1 \mathmr{and} \left(1 , 0\right)$ or

x intercept is at $x = 1 \mathmr{and} \left(1 , 0\right)$

y-intercept is found by putting $x = 0$ in the equation

$y = {\left(x - 1\right)}^{2} \mathmr{and} y = {\left(0 - 1\right)}^{2} \therefore y = 1 \mathmr{and} \left(0 , 1\right)$

y intercept is at $y = 1 \mathmr{and} \left(0 , 1\right)$

Additional points: Let $x = 3 \therefore y = {\left(3 - 1\right)}^{2} = 4 \mathmr{and} \left(3 , 4\right)$ and

$x = - 1 \therefore y = {\left(- 1 - 1\right)}^{2} = 4 \mathmr{and} \left(- 1 , 4\right)$

graph{x^2-2 x +1 [-10, 10, -5, 5]} [Ans]