The question asks for the number of intersection points between y = x² - 2x - 3 and y = -x + 1, with the graphs already drawn showing visible intersections. This system combines a quadratic and linear equation, solved by substitution. Setting equal: x² - 2x - 3 = -x + 1, which becomes x² - x - 4 = 0. The discriminant is b² - 4ac = (-1)² - 4(1)(-4) = 1 + 16 = 17 > 0, confirming two real solutions. Since the discriminant is positive, the parabola and line intersect at exactly 2 points. When reading graphs on the SAT, trust the algebraic verification over visual estimation.

We need to find how many times the parabola y = x² + 4x + 5 intersects the horizontal line y = 1. This system has a quadratic and a constant linear equation, solved by substitution. Setting equal: x² + 4x + 5 = 1, which simplifies to x² + 4x + 4 = 0. This factors as (x + 2)² = 0, giving x = -2 as a repeated root. Since we have one repeated solution, the line is tangent to the parabola at exactly one point. A common misconception is that a quadratic equation always has two distinct solutions, but when the discriminant equals zero, there's exactly one solution (the vertex touches the line).

We need to find how many times the parabola y = x² + 1 intersects the line y = -2x - 3. This system combines a quadratic and linear equation, solved by substitution. Setting equal: x² + 1 = -2x - 3, which gives x² + 2x + 4 = 0. To determine the number of real solutions, we check the discriminant: b² - 4ac = (2)² - 4(1)(4) = 4 - 16 = -12 < 0. Since the discriminant is negative, there are no real solutions - the parabola and line don't intersect. A common error is assuming every parabola-line pair must intersect, but when the line is positioned entirely below or above the parabola, there are zero intersection points.

To find where the parabola y = -x² + 4x meets the line y = x, we solve this system using substitution. Setting equal: -x² + 4x = x, which simplifies to -x² + 3x = 0, then -x(x - 3) = 0. This gives x = 0 or x = 3. When x = 0, y = 0; when x = 3, y = 3. The solutions are (0,0) and (3,3). A common error is forgetting that x = 0 is a valid solution when factoring out x - always include it when you factor out a variable.

We need to find where the profit curve y = -x² + 10x - 21 meets the break-even line y = x + 3, then find the sum of the x-coordinates. This system combines a quadratic with a linear equation, solved by substitution. Setting equal: -x² + 10x - 21 = x + 3, which simplifies to -x² + 9x - 24 = 0, or x² - 9x + 24 = 0. For a quadratic ax² + bx + c = 0, the sum of roots equals -b/a, so x₁ + x₂ = -(-9)/1 = 9. A common error is trying to find individual roots when only their sum is needed - using Vieta's formulas saves time on the SAT.

We need to find when the ball and drone are at the same height, which occurs where the parabola y = -x² + 6x + 1 intersects the line y = 2x + 1. This is a system with one quadratic and one linear equation, so we'll use substitution. Setting the equations equal: -x² + 6x + 1 = 2x + 1, which simplifies to -x² + 4x = 0, then -x(x - 4) = 0, giving x = 0 or x = 4. When x = 0, y = 2(0) + 1 = 1; when x = 4, y = 2(4) + 1 = 9. A common error is forgetting to find the y-values after solving for x, but the question asks for complete coordinate pairs. Since a parabola and line can intersect at most twice, finding two solutions confirms we have all intersection points.

We need to find where the parabola y = -x² + 4x + 4 intersects the horizontal line y = 4. This system has a quadratic and a constant linear equation, solved by substitution. Setting equal: -x² + 4x + 4 = 4, which simplifies to -x² + 4x = 0, then -x(x - 4) = 0. This gives x = 0 or x = 4. Since y = 4 for both solutions (given by the line equation), our intersection points are (0,4) and (4,4). A common error is thinking a horizontal line can only intersect a parabola once, but it can intersect 0, 1, or 2 times depending on the parabola's position and shape.

The question asks for the number of intersection points between the parabola y = x² - 4x + 1 and the line y = 3x - 5. This is a system of one quadratic and one linear equation, which we solve by substitution. Setting equal: x² - 4x + 1 = 3x - 5, which becomes x² - 7x + 6 = 0. Using the discriminant b² - 4ac = (-7)² - 4(1)(6) = 49 - 24 = 25 > 0, we know there are two real solutions. The positive discriminant tells us the parabola and line intersect at exactly 2 points. When solving systems graphically or algebraically, remember that a parabola and line can intersect 0, 1, or 2 times depending on their relative positions.

To find where the line y = 3x + 2 meets the parabola y = x² + 2x + 2, we solve this system using substitution. Setting equal: 3x + 2 = x² + 2x + 2, which simplifies to 0 = x² - x, then 0 = x(x - 1). This gives x = 0 or x = 1. When x = 0, y = 3(0) + 2 = 2; when x = 1, y = 3(1) + 2 = 5. The solutions are (0,2) and (1,5). A common error is making arithmetic mistakes when collecting like terms - always double-check by substituting solutions back into both original equations.

We need to find how many times the parabola y = x² - 9 reaches the height y = 6. This system has a quadratic and a constant linear equation, solved by substitution. Setting equal: x² - 9 = 6, which gives x² = 15. This has two real solutions: x = ±√15. Since we get two distinct x-values, the horizontal line y = 6 intersects the parabola at exactly 2 points. A key insight: when a horizontal line intersects a parabola above its vertex, there are always two intersection points.