Find the x and y Intercepts y=x^2-2x+1

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

Step 1

Find the x-intercepts.

To find the x-intercept(s), substitute in $0$ for $y$ and solve for $x$ .

$0 = x^{2} - 2 x + 1$

Solve the equation.

Rewrite the equation as $x^{2} - 2 x + 1 = 0$ .

$x^{2} - 2 x + 1 = 0$

Factor using the perfect square rule.

Rewrite $1$ as $1^{2}$ .

$x^{2} - 2 x + 1^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 x = 2 \cdot x \cdot 1$

Rewrite the polynomial.

$x^{2} - 2 \cdot x \cdot 1 + 1^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 1$ .

${(x - 1)}^{2} = 0$

Set the $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

x-intercept(s) in point form.

x-intercept(s): $(1, 0)$

Step 2

Find the y-intercepts.

To find the y-intercept(s), substitute in $0$ for $x$ and solve for $y$ .

$y = {(0)}^{2} - 2 \cdot 0 + 1$

Solve the equation.

Remove parentheses.

$y = 0^{2} - 2 \cdot 0 + 1$

Remove parentheses.

$y = {(0)}^{2} - 2 \cdot 0 + 1$

Simplify ${(0)}^{2} - 2 \cdot 0 + 1$ .

Simplify each term.

Raising $0$ to any positive power yields $0$ .

$y = 0 - 2 \cdot 0 + 1$

Multiply $- 2$ by $0$ .

$y = 0 + 0 + 1$

Simplify by adding numbers.

Add $0$ and $0$ .

$y = 0 + 1$

Add $0$ and $1$ .

$y = 1$

y-intercept(s) in point form.

y-intercept(s): $(0, 1)$

Step 3

List the intersections.

x-intercept(s): $(1, 0)$

y-intercept(s): $(0, 1)$

Step 4

Find the x and y Intercepts y=x^2-2x+1

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