Question

determine whether the graph of the equation is symmetric with respect to the x - axis, y - axis, origin, or none of these.
y = x² + 2x + 1
your answer
part 1 of 3
test for symmetry with respect to the y - axis. replace x by -x. write the equation in expanded form.
y = x² + 2x + 1
y = (-x)² + 2(-x) + 1
y = x² - 2x + 1
this equation is not equivalent to the original equation. thus, the graph of the equation is not symmetric with respect to the y - axis.
part: 1 / 3
part 2 of 3
test for symmetry with respect to the x - axis. replace y by -y. write the equation in expanded form.
y = x² + 2x + 1

• y = x² + 2x + 1

y = □
this equation is select to the original equation. thus, the graph of the equation select symmetric with respect to the x - axis.

Explanation:

Step1: Start with the equation after replacing \( y \) with \( -y \)

We have \( -y = x^2 + 2x + 1 \). To solve for \( y \), we multiply both sides of the equation by -1.

Step2: Multiply both sides by -1

Multiplying each term on the right - hand side by -1, we get \( y=-x^{2}-2x - 1 \).

Now, we compare this equation \( y=-x^{2}-2x - 1 \) with the original equation \( y = x^{2}+2x + 1 \). These two equations are not equivalent because the coefficients of \( x^{2}\), \( x \), and the constant term are different (except for the sign of the coefficients). So, the equation \( y=-x^{2}-2x - 1 \) is not equivalent to the original equation \( y = x^{2}+2x + 1 \), and thus the graph of the equation is not symmetric with respect to the \( x \)-axis.

Answer:

For the first blank (the equation for \( y \) after solving \( -y=x^{2}+2x + 1 \) for \( y \)): \( -x^{2}-2x - 1 \)

For the "Select" (whether the equation is equivalent) : not equivalent

For the "Select" (whether symmetric with respect to \( x \)-axis): is not