This question asks for the y-intercept of a downward-opening parabola with vertex at (0, 4) and x-intercepts at (-2, 0) and (2, 0). The graph shows a symmetric curve peaking at (0, 4) and crossing the x-axis at ±2, with the y-axis as the axis of symmetry, indicating the vertex is on the y-axis. The y-intercept is the y-value at x=0, which is directly the vertex's y-coordinate, 4. This connects the visual peak and symmetry to the intercept, as the parabola's equation like y = $-x^2$ + 4 would confirm y(0)=4. A key error is confusing x-intercepts like 2 with the y-intercept, or misidentifying the vertex height. When finding intercepts from graphs, distinguish between x=0 for y-intercept and y=0 for x-intercepts, especially in symmetric figures.

This question seeks the equation of a decreasing exponential curve passing through (0, 4) and (1, 2) with a horizontal asymptote at y=0. The graph depicts a smooth decay approaching the x-axis from above, starting at y=4 when x=0 and halving to y=2 at x=1, indicating a base between 0 and 1 without vertical shifts. The general form y = a * $b^x$ fits, where plugging x=0 gives a=4, and x=1 gives 4b=2 so b=1/2, yielding $y=4*(1/2)^x$. This connects the visual halving and y-intercept to the exponential parameters, confirming no asymptote shift as it approaches y=0. A key error is adding a vertical shift like +1, mistaking the asymptote, or choosing an increasing base like 2. When matching exponentials, use two points to solve for base and coefficient, and check the asymptote to distinguish growth from decay.

The question seeks the equation for a graph with a sharp corner at the origin and rays upward to (2, 2) and (-2, 2), symmetric about the y-axis. The graph resembles a V-shape opening upward, with equal slopes of 1 on both sides, indicating an absolute value function centered at the origin. The form y = |x| fits, as at x=2, y=2, and at x=-2, y=2, matching the points and symmetry. This links the visual corner and rays to the absolute value's algebraic property of reflecting negative inputs. A common mistake is choosing a linear equation like y=x, which doesn't account for the left ray's upward direction. For piecewise-like graphs, identify symmetry and corners to select absolute value over linear functions.

The question requires the equation of an exponential growth curve with asymptote y=0, passing through (0, 1) and (2, 9). The graph shows a curve rising from (0, 1) and steepening to (2, 9), approaching the x-axis leftward, typical of base greater than 1 without shifts. The form y = a * $b^x$ fits; at x=0, a=1, and at x=2, 1 * $b^2$ =9 so b=3, giving $y=3^x$. This links the y-intercept and growth rate to the base, verifying $3^2$=9 matches the point. A common error is assuming base 2 from misreading (2, 9) as (2, 8) or 4, or adding a shift. For exponentials, use the y-intercept for a and another point for b, checking the asymptote for no vertical shift.

This question requires the equation of a rational function with vertical asymptote at x=2, horizontal at y=1, and passing through (3, 2). The graph shows a hyperbolic curve approaching x=2 vertically and y=1 horizontally, with one branch passing through (3, 2) to the right of the asymptote. The form y = a/(x - h) + k places the vertical asymptote at x=h and horizontal at y=k, so h=2 and k=1; plugging (3, 2) gives 2 = a/(3-2) + 1, so a=1, yielding y=1/(x-2)+1. This connects the visual asymptotes and point to the reciprocal structure shifted appropriately. A key error is changing the sign of the shift, like x+2, altering the vertical asymptote, or adjusting the numerator incorrectly. When identifying rationals, match asymptotes first, then use a point to find the coefficient, ensuring the graph's behavior aligns.

This question asks for the equation of a parabola given its graph features, including vertex and x-intercepts. The graph shows an upward-opening parabola with vertex at (1, -4), x-intercepts at (-1, 0) and (3, 0), and symmetry about x=1, indicating a standard quadratic shape shifted right and down. To find the matching equation, recognize that the vertex form y = (x - $h)^2$ + k directly incorporates the vertex (h, k), so for (1, -4), it suggests y = (x - $1)^2$ - 4. Verifying the x-intercepts by setting y = 0 gives (x - $1)^2$ = 4, so x = 1 ± 2, matching -1 and 3, confirming the algebraic properties align with the visual roots and minimum point. A key error is misreading the vertex as (-1, -4), leading to choice D, or ignoring the vertex form and mistakenly using the factored form without verification. When analyzing graphs, always connect key points like vertices and intercepts to both forms of quadratic equations to ensure consistency.

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.

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 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 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.