The question asks for the domain of a line segment from (-4,2) to (4,2) with closed endpoints. The domain consists of all possible x-values for which the function is defined. Looking at the horizontal line segment, x ranges from -4 to 4, and since the endpoints are closed (filled in), we include both -4 and 4. Therefore, the domain is [-4,4], using square brackets to indicate inclusion of endpoints. A common error is giving the range instead (which would be just {2}) or using parentheses (-4,4) which would exclude the endpoints. For graphed functions, scan left to right to identify the domain, and use brackets for closed endpoints and parentheses for open endpoints.

We need to find the equation of a parabola opening downward with x-intercepts at x = -1 and x = 3, crossing the y-axis at y = 3. Since the parabola opens downward, the leading coefficient must be negative. The factored form with x-intercepts at -1 and 3 is y = a(x+1)(x-3) where a < 0. To find a, we use the y-intercept: when x = 0, y = 3. Substituting: 3 = a(0+1)(0-3) = a(1)(-3) = -3a, so a = -1. Therefore, y = -(x+1)(x-3), which matches option B. Common errors include forgetting the negative sign for downward-opening parabolas or confusing the signs in the factors. When a parabola crosses the x-axis at x = c, the corresponding factor is (x-c).

The question describes a V-shaped graph (absolute value function) with vertex at (2,-1) and slopes ±1. The general form of an absolute value function is f(x) = a|x-h| + k where (h,k) is the vertex. With vertex at (2,-1), we have h = 2 and k = -1, giving f(x) = a|x-2| - 1. Since the slopes are ±1, we know a = 1 (if a were negative, the V would open downward). Therefore, f(x) = |x-2| - 1, which matches option A. Common mistakes include using |x+2| instead of |x-2| (remember that |x-h| shifts right by h units) or adding 1 instead of subtracting. To verify transformations, always check that the vertex location matches your equation.

We need to identify the rational function with vertical asymptote at x = 2, horizontal asymptote at y = 1, passing through (3,2). A vertical asymptote at x = 2 means the denominator has a factor (x-2). A horizontal asymptote at y = 1 suggests the function approaches 1 as x approaches infinity, indicating a form like y = 1 + k/(x-2). Using the point (3,2): 2 = 1 + k/(3-2) = 1 + k/1, so k = 1. Therefore, y = 1 + 1/(x-2), which matches option A. Common errors include using (x+2) in the denominator (which would put the asymptote at x = -2) or forgetting to account for the horizontal asymptote. When working with rational functions, identify asymptotes first to determine the basic structure.

We need to find the equation of a decreasing exponential curve passing through (0,4) and (2,1) that approaches y = 0 as x increases. For a decreasing exponential y = $a·b^x$ where 0 < b < 1, using point (0,4): 4 = $a·b^0$ = a, so a = 4. Using point (2,1): 1 = $4·b^2$, so $b^2$ = 1/4, giving b = 1/2. Therefore, y = $4(1/2)^x$, which matches option A. Let's verify: at x = 0, y = $4(1/2)^0$ = 4·1 = 4 ✓; at x = 2, y = $4(1/2)^2$ = 4·(1/4) = 1 ✓. Common mistakes include using an increasing base like 2 or forgetting the initial value coefficient. For decreasing exponentials, the base must be between 0 and 1.

The question asks for the y-intercept of a line passing through (0,-2) and (3,4). The y-intercept is the y-coordinate where the line crosses the y-axis, which occurs when x = 0. Looking at the given points, we see that (0,-2) is already on the y-axis, so the y-intercept is -2. This can be verified by finding the line's equation: slope = (4-(-2))/(3-0) = 6/3 = 2, giving y = 2x - 2, which confirms the y-intercept is -2. A common error is confusing the y-intercept with the x-intercept or misreading the coordinates. Remember that the y-intercept is always the point where x = 0.

We need to find the equation of a square root curve starting at (1,-2) and passing through (5,0). The general form is y = a√(x-h) + k where (h,k) is the starting point. With starting point (1,-2), we have h = 1 and k = -2, giving y = a√(x-1) - 2. Using point (5,0): 0 = a√(5-1) - 2 = a√4 - 2 = 2a - 2, so a = 1. Therefore, y = √(x-1) - 2, which matches option A. Common errors include using (x+1) instead of (x-1) inside the radical (remember that √(x-h) shifts right by h) or adding 2 instead of subtracting. For square root functions, the domain starts at x = h, and vertical shifts affect the range.

The question asks for the axis of symmetry of a parabola with x-intercepts at x = -2 and x = 4, and vertex at (1,-9). The axis of symmetry is a vertical line passing through the vertex, so it has equation x = h where h is the x-coordinate of the vertex. Since the vertex is at (1,-9), the axis of symmetry is x = 1. This can be verified by noting that the axis of symmetry is equidistant from both x-intercepts: the midpoint of -2 and 4 is (-2+4)/2 = 2/2 = 1. A common error is using the midpoint formula incorrectly or confusing the axis of symmetry with one of the x-intercepts. For any parabola, the axis of symmetry passes through the vertex.

The question asks whether a piecewise linear graph with two segments meeting at (0,3) represents a function. The graph consists of one segment from (-4,-1) to (0,3) and another from (0,3) to (4,-1), with (0,3) included. To determine if this is a function, we apply the vertical line test: every vertical line must intersect the graph at most once. Since the two segments meet at exactly one point (0,3) and don't overlap elsewhere, each x-value corresponds to exactly one y-value. Therefore, the graph passes the vertical line test and is a function. A common concern is the meeting point, but since both segments include the same point (0,3), there's no ambiguity. When checking if a graph is a function, look for any x-value that might have multiple y-values.

The question asks for the range of a horizontal line at y = -3. The range of a function consists of all possible y-values (outputs). For a horizontal line at y = -3, every point on the line has y-coordinate -3, regardless of the x-value. Therefore, the range contains only one value: -3. This is written as the set {-3}, not as an interval. A common mistake is giving the domain (which would be (-∞,∞)) or writing the range as an interval like [-3,-3]. Remember that for horizontal lines, the range is a single value, while the domain is all real numbers. When a function outputs only one value, use set notation with curly braces.