Find the x and y Intercepts f(x)=(x^2+2x+1)/(2x+2)

$f (x) = \frac{x^{2} + 2 x + 1}{2 x + 2}$

Find the x-intercepts.

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

$0 = \frac{x^{2} + 2 x + 1}{2 x + 2}$

Solve the equation.

Set the numerator equal to zero.

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

Solve the equation for $x$ .

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 $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Exclude the solutions that do not make $0 = \frac{x^{2} + 2 x + 1}{2 x + 2}$ true.

$No solution$

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

x-intercept(s): $None$

Find the y-intercepts.

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

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

Solve the equation.

Remove parentheses.

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

Remove parentheses.

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

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

Simplify the numerator.

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

$y = \frac{0 + 2 (0) + 1}{2 (0) + 2}$

Multiply $2$ by $0$ .

$y = \frac{0 + 0 + 1}{2 (0) + 2}$

Add $0$ and $0$ .

$y = \frac{0 + 1}{2 (0) + 2}$

Add $0$ and $1$ .

$y = \frac{1}{2 (0) + 2}$

Simplify the denominator.

Multiply $2$ by $0$ .

$y = \frac{1}{0 + 2}$

Add $0$ and $2$ .

$y = \frac{1}{2}$

y-intercept(s) in point form.

y-intercept(s): $(0, \frac{1}{2})$

List the intersections.

x-intercept(s): $None$

y-intercept(s): $(0, \frac{1}{2})$

Find the x and y Intercepts f(x)=(x^2+2x+1)/(2x+2)

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