1. We are asked to find where the function $f(x) = \frac{2x^2 - 6x + 4}{-x^2 - 3x + 10}$ has a jump (discontinuity) and to give the corresponding vertical asymptotes (VA). 2. First, recall that vertical asymptotes occur where the denominator is zero and the numerator is not zero at those points. 3. Find the zeros of the denominator by solving: $$-x^2 - 3x + 10 = 0$$ Multiply both sides by $-1$ to simplify: $$\cancel{-1} \cdot (-x^2 - 3x + 10) = \cancel{-1} \cdot 0 \Rightarrow x^2 + 3x - 10 = 0$$ 4. Factor the quadratic: $$x^2 + 3x - 10 = (x + 5)(x - 2) = 0$$ So the denominator is zero at $x = -5$ and $x = 2$. 5. Check the numerator at these points: $$2x^2 - 6x + 4$$ At $x = -5$: $$2(-5)^2 - 6(-5) + 4 = 2(25) + 30 + 4 = 50 + 30 + 4 = 84 \neq 0$$ At $x = 2$: $$2(2)^2 - 6(2) + 4 = 2(4) - 12 + 4 = 8 - 12 + 4 = 0$$ 6. Since the numerator is zero at $x=2$, we check if the factor $(x-2)$ cancels with the denominator: Factor numerator: $$2x^2 - 6x + 4 = 2(x^2 - 3x + 2) = 2(x - 1)(x - 2)$$ Denominator: $$-x^2 - 3x + 10 = -(x^2 + 3x - 10) = -(x + 5)(x - 2)$$ 7. Cancel the common factor $(x - 2)$: $$f(x) = \frac{2(x - 1)\cancel{(x - 2)}}{-\cancel{(x - 2)}(x + 5)} = \frac{2(x - 1)}{-(x + 5)} = -\frac{2(x - 1)}{x + 5}$$ 8. The factor $(x - 2)$ cancels, so at $x=2$ there is a removable discontinuity (a hole), not a vertical asymptote. 9. At $x = -5$, the denominator is zero and numerator is not zero, so there is a vertical asymptote (VA) at $x = -5$. 10. Summary: - Vertical asymptote at $x = -5$ - Removable discontinuity (hole) at $x = 2$ Final answer: The function has a vertical asymptote at $x = -5$ and a removable discontinuity (hole) at $x = 2$.