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hw 2.7 analyzing graphs and piecewise-defined functions question 6 of 1…

Question

hw 2.7 analyzing graphs and piecewise-defined functions
question 6 of 11 (2 points) | question attempt: 1 of unlimited
part: 2 / 3
part 3 of 3
test for symmetry with respect to the origin. replace x by -x and y by -y. write the equation in expanded form.
( y = x^2 + 8x + 3 )
( -y = (-x)^2 + 8(-x) + 3 )
( -y = x^2 - 8x + 3 )
( y = x^2 - 8x - 3 )
this equation is not equivalent to the original equation. thus, the graph of the equation is not symmetric with respect to the origin.
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Explanation:

Step 1: Replace \( x \) with \( -x \) and \( y \) with \( -y \)

Given the original equation \( y = x^2 + 8x + 3 \), we substitute \( x \) with \( -x \) and \( y \) with \( -y \):
\[
-y = (-x)^2 + 8(-x) + 3
\]

Step 2: Simplify the right - hand side

We know that \( (-x)^2=x^2 \) and \( 8(-x)=-8x \). So the equation becomes:
\[
-y = x^2-8x + 3
\]

Step 3: Solve for \( y \)

Multiply both sides of the equation \( -y=x^2 - 8x+3 \) by \( - 1 \) to solve for \( y \). When we multiply each term on the right - hand side by \( -1 \), we get:
\[
y=-x^2 + 8x-3
\]
Now, we compare this equation \( y=-x^2 + 8x - 3 \) with the original equation \( y=x^2 + 8x+3 \). Since the coefficients of \( x^2 \) are different (\( - 1\) vs \( 1\)) and the constant terms are different (\( - 3\) vs \( 3\)), the two equations are not equivalent. So the graph of the equation is not symmetric with respect to the origin.

Answer:

When we replace \( x \) with \( -x \) and \( y \) with \( -y \) in \( y = x^2+8x + 3 \), we get \( -y=(-x)^2+8(-x)+3\), which simplifies to \( -y=x^2 - 8x + 3 \), and then \( y=-x^2 + 8x-3 \). This equation is not equivalent to the original equation \( y=x^2 + 8x + 3 \), so the graph of the equation is not symmetric with respect to the origin. The correct equation after solving for \( y \) is \( y=-x^2 + 8x - 3 \).