1. **State the problem:** We want to show mathematically that the line $y=5x+10$ intersects only one branch of the graph of $f(x)=3|x-2|-10$. 2. **Recall the definition of $f(x)$:** The function $f(x)=3|x-2|-10$ is a V-shaped graph with vertex at $P(2,-10)$. 3. **Rewrite $f(x)$ as a piecewise function:** $$ f(x) = \begin{cases} 3(x-2)-10 = 3x-6-10 = 3x-16 & \text{if } x \geq 2 \\ -3(x-2)-10 = -3x+6-10 = -3x-4 & \text{if } x < 2 \end{cases} $$ 4. **Set the line equal to each piece to find intersections:** - For $x \geq 2$: $$ 3x - 16 = 5x + 10 $$ Rearranging: $$ 3x - 16 = 5x + 10 \implies 3x - 5x = 10 + 16 \implies -2x = 26 \implies x = -13 $$ Since $x=-13$ is not $\geq 2$, this solution is invalid for this branch. - For $x < 2$: $$ -3x - 4 = 5x + 10 $$ Rearranging: $$ -3x - 4 = 5x + 10 \implies -3x - 5x = 10 + 4 \implies -8x = 14 \implies x = -\frac{14}{8} = -\frac{7}{4} = -1.75 $$ Since $x = -1.75 < 2$, this solution is valid for this branch. 5. **Conclusion:** The line $y=5x+10$ intersects the graph of $f(x)$ only once, at $x = -1.75$, which lies on the left branch ($x<2$). There is no intersection on the right branch ($x \geq 2$). Hence, the line intersects only one branch of the modulus graph. **Final answer:** The line $y=5x+10$ intersects the graph of $f(x)=3|x-2|-10$ only once, at $x=-\frac{7}{4}$ on the left branch.