1. The problem is to find the range of values of $k$ such that the graphs of $y = |(x+1)(x-3)|$ and $y = x + k$ have exactly two common points. 2. The absolute value function $y = |(x+1)(x-3)|$ can be split into two cases based on the sign of $(x+1)(x-3)$: - For $x \leq -1$ or $x \geq 3$, $(x+1)(x-3) \geq 0$, so $y = (x+1)(x-3)$. - For $-1 < x < 3$, $(x+1)(x-3) < 0$, so $y = -(x+1)(x-3)$. 3. The graph of $y = |(x+1)(x-3)|$ is thus composed of two parabolas: - An upward-opening parabola $y = (x+1)(x-3)$ outside the interval $[-1,3]$. - A downward-opening parabola $y = -(x+1)(x-3)$ inside the interval $(-1,3)$. 4. The line $y = x + k$ intersects these parabolas. To have exactly two common points, the line must intersect the graph in a way that it touches or crosses the parabolas at exactly two points. 5. Consider the intersection with the downward parabola $y = -(x+1)(x-3)$ for $-1 < x < 3$: Set $-(x+1)(x-3) = x + k$. Expanding: $$-(x^2 - 2x - 3) = x + k \implies -x^2 + 2x + 3 = x + k$$ Rearranged: $$-x^2 + 2x + 3 - x - k = 0 \implies -x^2 + x + 3 - k = 0$$ Multiply both sides by $-1$: $$x^2 - x + k - 3 = 0$$ 6. For the quadratic equation $x^2 - x + (k - 3) = 0$ to have exactly one solution (tangency), the discriminant $D$ must be zero: $$D = (-1)^2 - 4 \times 1 \times (k - 3) = 1 - 4(k - 3) = 0$$ Solve for $k$: $$1 - 4k + 12 = 0 \implies 13 - 4k = 0 \implies k = \frac{13}{4} = 3.25$$ 7. Next, find where the line $y = x + k$ passes through the points where the absolute value function changes definition: - At $x = -1$, $y = 0$ for the absolute value function, so: $$0 = -1 + k \implies k = 1$$ - At $x = 3$, $y = 0$ for the absolute value function, so: $$0 = 3 + k \implies k = -3$$ 8. From the graph and these values: - For $k$ between $-3$ and $1$, the line intersects the graph exactly twice. - For $k > 3.25$, the line also intersects the graph exactly twice. 9. Therefore, the range of $k$ for which the graphs have exactly two common points is: $$-3 < k < 1 \quad \text{or} \quad k > \frac{13}{4}$$ This explains how the graph is constructed and how the values of $k$ are determined based on the intersection points and tangency conditions.