Find the x and y Intercepts y=(x+1)^2-7(x+1)+12

$y = {(x + 1)}^{2} - 7 (x + 1) + 12$

Find the x-intercepts.

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

$0 = {(x + 1)}^{2} - 7 (x + 1) + 12$

Solve the equation.

Rewrite the equation as ${(x + 1)}^{2} - 7 (x + 1) + 12 = 0$ .

${(x + 1)}^{2} - 7 (x + 1) + 12 = 0$

Factor the left side of the equation.

Let $u = x + 1$ . Substitute $u$ for all occurrences of $x + 1$ .

$u^{2} - 7 u + 12 = 0$

Factor $u^{2} - 7 u + 12$ 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 $12$ and whose sum is $- 7$ .

$- 4, - 3$

Write the factored form using these integers.

$(u - 4) (u - 3) = 0$

Replace all occurrences of $u$ with $x + 1$ .

$(x + 1 - 4) (x + 1 - 3) = 0$

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

$x + 1 - 4 = 0$

$x + 1 - 3 = 0$

Set $x + 1 - 4$ equal to $0$ and solve for $x$ .

Set $x + 1 - 4$ equal to $0$ .

$x + 1 - 4 = 0$

Solve $x + 1 - 4 = 0$ for $x$ .

Subtract $4$ from $1$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Set $x + 1 - 3$ equal to $0$ and solve for $x$ .

Set $x + 1 - 3$ equal to $0$ .

$x + 1 - 3 = 0$

Solve $x + 1 - 3 = 0$ for $x$ .

Subtract $3$ from $1$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

The final solution is all the values that make $(x + 1 - 4) (x + 1 - 3) = 0$ true.

$x = 3, 2$

x-intercept(s) in point form.

x-intercept(s): $(3, 0), (2, 0)$

Find the y-intercepts.

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

$y = {((0) + 1)}^{2} - 7 ((0) + 1) + 12$

Solve the equation.

Remove parentheses.

$y = {(0 + 1)}^{2} - 7 ((0) + 1) + 12$

Remove parentheses.

$y = {(0 + 1)}^{2} - 7 (0 + 1) + 12$

Remove parentheses.

$y = {((0) + 1)}^{2} - 7 ((0) + 1) + 12$

Simplify ${((0) + 1)}^{2} - 7 ((0) + 1) + 12$ .

Simplify each term.

Add $0$ and $1$ .

$y = 1^{2} - 7 ((0) + 1) + 12$

One to any power is one.

$y = 1 - 7 ((0) + 1) + 12$

Add $0$ and $1$ .

$y = 1 - 7 \cdot 1 + 12$

Multiply $- 7$ by $1$ .

$y = 1 - 7 + 12$

Simplify by adding and subtracting.

Subtract $7$ from $1$ .

$y = - 6 + 12$

Add $- 6$ and $12$ .

$y = 6$

y-intercept(s) in point form.

y-intercept(s): $(0, 6)$

List the intersections.

x-intercept(s): $(3, 0), (2, 0)$

y-intercept(s): $(0, 6)$

Find the x and y Intercepts y=(x+1)^2-7(x+1)+12

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