Question

Difficulty: HardQuadratic Equations

A parabola in the xyxy-plane has equation y=a(xh)2+ky = a(x - h)^2 + k, where aa, hh, and kk are constants. The parabola passes through the points (2,13)(2, 13) and (8,13)(8, 13) in the xyxy-plane. If the minimum value of the quadratic function defined by this equation is 5-5, what is the value of aa?

Answer: 2

Answer

The correct answer is 2.
The correct answer is 2. The axis of symmetry of the parabola is halfway between the x-coordinates of the two symmetric points (2,13)(2, 13) and (8,13)(8, 13), which is x=5x = 5. Since the minimum value of the function is 5-5, the vertex of the parabola is (5,5)(5, -5). Writing the equation in vertex form, y=a(x5)25y = a(x - 5)^2 - 5, and substituting the point (2,13)(2, 13) yields 13=a(25)2513 = a(2 - 5)^2 - 5, which simplifies to 18=9a18 = 9a, so a=2a = 2.

Step-by-Step Solution

1
Find the axis of symmetry x=hx = h using the symmetric points (2,13)(2, 13) and (8,13)(8, 13).
h=5h = 5
Since the points (2,13)(2, 13) and (8,13)(8, 13) have the same yy-coordinate, the axis of symmetry of the parabola must lie exactly halfway between their xx-coordinates: h=2+82=5h = \frac{2 + 8}{2} = 5.
2
Determine the vertex (h,k)(h, k) of the parabola.
(5,5)(5, -5)
The minimum value of a quadratic function occurs at its vertex. The given minimum value is 5-5, which represents the yy-coordinate (kk) of the vertex. Combining this with h=5h = 5, the vertex is (5,5)(5, -5).
3
Write the equation of the parabola in vertex form.
y=a(x5)25y = a(x - 5)^2 - 5
The vertex form of a quadratic equation is y=a(xh)2+ky = a(x - h)^2 + k. Substituting the values h=5h = 5 and k=5k = -5 yields this equation.
4
Substitute one of the given points into the equation to solve for aa.
a=2a = 2
Substituting (2,13)(2, 13) into the equation gives 13=a(25)2518=9aa=213 = a(2 - 5)^2 - 5 \Rightarrow 18 = 9a \Rightarrow a = 2.

Key Concept

Finding the equation of a parabola using symmetry and vertex form
Estimated Time:2m 0s
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