https://file.aiquickdraw.com/mathbot/file/1726524190243oswevrmk.jpg

The equation given is:

\[
x^2 + 8x + 3 = |x + 1| + 4
\]

To solve this using graphing, we typically compare the left-hand side (a quadratic function) with the right-hand side (a piecewise function due to the absolute value).

1. **Quadratic**: The left-hand side is \( y = x^2 + 8x + 3 \).
2. **Absolute Value**: The right-hand side is \( y = |x + 1| + 4 \).

We need to find the intersections of these two functions, which would give the solutions to the equation.

### Looking at the graphs:
- The graph marked as **B** shows two curves, one representing a quadratic and the other representing a piecewise linear function (due to the absolute value).
- The intersections of these curves represent the solutions to the equation.

From graph **B**, it appears that there are two intersection points, one slightly left of the y-axis and another to the right.

### Approximating the solutions:
The intersections seem to occur around:
- \( x \approx -7.2 \)
- \( x \approx -0.5 \)

So, the approximate solutions are:
\[
x \approx -7.2, -0.5
\]

Let me know if you'd like further details!

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Here are 5 related questions:
1. How does the absolute value affect the graph of the function?
2. What is the significance of the intersection points on these graphs?
3. How can you verify these solutions algebraically?
4. How would the graph change if the quadratic term was different?
5. How do you graph absolute value equations by hand?

**Tip:** When dealing with absolute value equations, always consider their piecewise nature. This makes graphing easier and helps in solving algebraically.

Use graphing technology to approximate the solutions of the equation x^2 + 8x + 3 = |x + 1| + 4.

Learn how to solve the quadratic equation x^2 + 8x + 3 = |x + 1| + 4 using graphing technology. This problem explores key algebraic concepts, including quadratic equations and absolute value, and is suitable for students in Grades 9-12. Approximate solutions are found graphically at x ≈ -7.2 and x ≈ -0.5.

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