Find the x and y Intercepts y=x^2-4x+3

$y = x^{2} - 4 x + 3$

Find the x-intercepts.

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

$0 = x^{2} - 4 x + 3$

Solve the equation.

Rewrite the equation as $x^{2} - 4 x + 3 = 0$ .

$x^{2} - 4 x + 3 = 0$

Factor $x^{2} - 4 x + 3$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $3$ and whose sum is $- 4$ .

$- 3, - 1$

Write the factored form using these integers.

$(x - 3) (x - 1) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 3 = 0$

$x - 1 = 0$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Set $x - 1$ equal to $0$ and solve for $x$ .

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

The final solution is all the values that make $(x - 3) (x - 1) = 0$ true.

$x = 3, 1$

x-intercept(s) in point form.

x-intercept(s): $(3, 0), (1, 0)$

Find the y-intercepts.

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

$y = {(0)}^{2} - 4 \cdot 0 + 3$

Solve the equation.

Remove parentheses.

$y = 0^{2} - 4 \cdot 0 + 3$

Remove parentheses.

$y = {(0)}^{2} - 4 \cdot 0 + 3$

Simplify ${(0)}^{2} - 4 \cdot 0 + 3$ .

Simplify each term.

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

$y = 0 - 4 \cdot 0 + 3$

Multiply $- 4$ by $0$ .

$y = 0 + 0 + 3$

Simplify by adding numbers.

Add $0$ and $0$ .

$y = 0 + 3$

Add $0$ and $3$ .

$y = 3$

y-intercept(s) in point form.

y-intercept(s): $(0, 3)$

List the intersections.

x-intercept(s): $(3, 0), (1, 0)$

y-intercept(s): $(0, 3)$

Find the x and y Intercepts y=x^2-4x+3

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