To solve this problem, we need to determine the positive $x$-value where the two equations intersect. The equations are:

1. $y = -x + 4$ (a linear equation)
2. $y = x^2 - 2x + 1$ (a quadratic equation)

Step 1: Set the equations equal to each other

Since both equations equal $y$, we can set them equal: $-x + 4 = x^2 - 2x + 1$

Step 2: Rearrange into standard quadratic form

Bring all terms to one side: $0 = x^2 - x - x + 1 - 4$ $x^2 - 2x - 3 = 0$

Step 3: Solve the quadratic equation

Factorize $x^2 - 2x - 3$: $x^2 - 2x - 3 = (x - 3)(x + 1)$

Set each factor to 0: $x - 3 = 0 \quad \Rightarrow \quad x = 3$ $x + 1 = 0 \quad \Rightarrow \quad x = -1$

Step 4: Identify the positive solution

The positive solution is $x = 3$.

Step 5: Verify if $x = 3$ is closest to one of the choices

Substitute $x = 3$ into the linear equation to find $y$: $y = -3 + 4 = 1$

Thus, the intersection point is $(3, 1)$, and the closest positive $x$-value is:

Answer: $4.0$ (Choice D)

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Would you like a step-by-step graph visualization? Or explanations about intersections and quadratics? 😊

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Related Questions:

1. How do you find the points of intersection of two curves?
2. What are the properties of quadratic equations and their graphs?
3. How does the discriminant help determine the nature of solutions to quadratics?
4. What are the advantages of factoring versus other methods of solving quadratics?
5. Can you find intersections using numerical methods or graphing tools?

Tip:

Always verify solutions by substituting back into the original equations to avoid errors!