Zeros Y Intercepts 9C5492

1. **State the problem:** We are given the function $$f(x) = \sqrt{x + 3} + 1$$ and need to find its zeros, y-intercept, and sketch its graph. 2. **Find the zeros of $f$:** Zeros occur where $$f(x) = 0$$. Set the function equal to zero: $$\sqrt{x + 3} + 1 = 0$$ Subtract 1 from both sides: $$\sqrt{x + 3} = -1$$ Since the square root function $$\sqrt{x + 3}$$ is always $$\geq 0$$ for its domain, it can never equal $$-1$$. Therefore, there are **no zeros** of the function. 3. **Find the y-intercept:** The y-intercept occurs when $$x = 0$$. Evaluate $$f(0)$$: $$f(0) = \sqrt{0 + 3} + 1 = \sqrt{3} + 1$$ So the y-intercept is at $$\left(0, \sqrt{3} + 1\right)$$. 4. **Domain of the function:** The expression inside the square root must be $$\geq 0$$: $$x + 3 \geq 0 \implies x \geq -3$$ So the domain is $$[-3, \infty)$$. 5. **Sketching the graph:** - The graph starts at $$x = -3$$. - At $$x = -3$$, $$f(-3) = \sqrt{-3 + 3} + 1 = 0 + 1 = 1$$. - The function increases as $$x$$ increases because $$\sqrt{x+3}$$ increases. - The graph is shifted up by 1 unit compared to $$y = \sqrt{x+3}$$. - The graph has no zeros and a y-intercept at $$\left(0, \sqrt{3} + 1\right)$$. **Final answer:** - No zeros. - Y-intercept at $$\left(0, \sqrt{3} + 1\right)$$. - Domain $$[-3, \infty)$$. - Increasing graph starting at point $$(-3,1)$$ and going upwards.