To determine which parabola intersects the line y = x + 5 at exactly one point, we need to consider the mathematical condition under which a parabola and a straight line share a single coordinate point. This situation occurs when the quadratic equation representing the parabola has a discriminant of zero.
The general form of a parabola can be expressed as:
y = ax² + bx + cTo find the parabola that intersects with the given line, we equate the two equations:
ax² + bx + c = x + 5Rearranging this gives us:
ax² + (b - 1)x + (c - 5) = 0This is a standard quadratic equation in the form of Ax² + Bx + C = 0, where:
- A = a
- B = b – 1
- C = c – 5
For the equation to have exactly one real solution, the discriminant (D) must equal zero:
D = B² - 4AC = 0Substituting our values, this becomes:
(b - 1)² - 4a(c - 5) = 0Now, by adjusting the coefficients (a, b, c), we can define various parabolas that adhere to this condition. For example:
- If we take a = 1, b = 1, and c = 6, the equation simplifies to:
(1 - 1)² - 4(1)(6 - 5) = 00 - 4(1)(1) = 0In conclusion, by selecting appropriate values for a, b, and c that satisfy the discriminant condition, you can create various parabolas intersecting the line y = x + 5 at just one point. If you’re looking for a specific instance, try y = x² + x + 6 or similar forms!