1. **Problem Statement:** We need to find the invariant points of the reciprocal of a linear function. An invariant point is where the function and its reciprocal intersect, meaning the point satisfies $f(x) = \frac{1}{f(x)}$. 2. **Formula and Explanation:** For a linear function $f(x) = mx + b$, the reciprocal function is $y = \frac{1}{mx + b}$. An invariant point satisfies: $$ f(x) = \frac{1}{f(x)} $$ which implies $$ mx + b = \frac{1}{mx + b} $$ 3. **Solve for invariant points:** Multiply both sides by $mx + b$: $$ (mx + b)(mx + b) = 1 $$ which is $$ (mx + b)^2 = 1 $$ 4. **Expand and simplify:** $$ (m^2 x^2 + 2mbx + b^2) = 1 $$ 5. **Rewrite as a quadratic equation:** $$ m^2 x^2 + 2mb x + (b^2 - 1) = 0 $$ 6. **Use the quadratic formula:** $$ x = \frac{-2mb \pm \sqrt{(2mb)^2 - 4 m^2 (b^2 - 1)}}{2 m^2} $$ Simplify inside the square root: $$ (2mb)^2 - 4 m^2 (b^2 - 1) = 4 m^2 b^2 - 4 m^2 b^2 + 4 m^2 = 4 m^2 $$ So, $$ x = \frac{-2mb \pm 2m}{2 m^2} = \frac{-mb \pm m}{m^2} = \frac{m(-b \pm 1)}{m^2} = \frac{-b \pm 1}{m} $$ 7. **Final invariant points:** $$ \left(\frac{-b + 1}{m}, 1 - b\right) \quad \text{and} \quad \left(\frac{-b - 1}{m}, -1 - b\right) $$ where the $y$-values come from substituting $x$ back into $f(x) = mx + b$. **Summary:** The reciprocal of a linear function $f(x) = mx + b$ has two invariant points at $$ \left(\frac{-b + 1}{m}, 1 - b\right) \quad \text{and} \quad \left(\frac{-b - 1}{m}, -1 - b\right) $$ These points lie on both the function and its reciprocal. This explains why different answers occur: the invariant points depend on the slope $m$ and intercept $b$ of the original linear function. q_count is 2 because the user asked about the reciprocal function sketch and invariant points, but we only solve the invariant points as per instructions.