1. **Problem Statement:** Find the domain, even or odd degree, leading coefficient, number of turning points, least possible degree, number of real zeroes, y-intercept, end behavior, and equation for the function $$f(x) = \sqrt{4+x}$$. 2. **Equation:** The function is given as $$f(x) = \sqrt{4+x} = (4+x)^{\frac{1}{2}}$$. 3. **Domain:** The expression inside the square root must be non-negative: $$4 + x \geq 0 \implies x \geq -4$$ So, the domain is $$[-4, \infty)$$. 4. **Degree:** Since the function is a square root (fractional exponent), it is not a polynomial, so it does not have a degree in the usual sense (degree is defined for polynomials only). 5. **Leading Coefficient:** Not applicable because this is not a polynomial. 6. **Number of Turning Points:** The function $$f(x) = \sqrt{4+x}$$ is increasing and concave down on its domain, so it has no turning points. 7. **Least Possible Degree:** Since this is not a polynomial, the concept of degree does not apply. 8. **Number of Real Zeroes:** Solve $$f(x) = 0$$: $$\sqrt{4+x} = 0 \implies 4 + x = 0 \implies x = -4$$ So, there is exactly one real zero at $$x = -4$$. 9. **Y-Intercept:** Evaluate $$f(0)$$: $$f(0) = \sqrt{4 + 0} = \sqrt{4} = 2$$ So, the y-intercept is at $$(0, 2)$$. 10. **End Behavior:** As $$x \to \infty$$, $$f(x) = \sqrt{4+x} \approx \sqrt{x} \to \infty$$. As $$x \to -4^+$$ (approaching -4 from the right), $$f(x) \to 0^+$$. 11. **Summary:** - Domain: $$[-4, \infty)$$ - Not a polynomial, so no degree or leading coefficient - Number of turning points: 0 - Number of real zeroes: 1 at $$x = -4$$ - Y-intercept: $$(0, 2)$$ - End behavior: $$f(x) \to \infty$$ as $$x \to \infty$$, and $$f(x) \to 0$$ as $$x \to -4^+$$ - Equation: $$f(x) = \sqrt{4+x}$$