| What's my equation? |
The graph shown below consists of a line segment connected to a piece of a parabola, with the intersection at (0, 2). The derivative at the point of intersection is continuous, and each tick mark on the axes represents one unit. Find the equation of the parabola in the form p(x) = ax2 + bx + c.
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Problem Moderated by: MrT |
| Problem Solution |
Since the parabola passes through the points (0, 2) and (2, 4), we arrive at the following two equations:
EQN 1: 2 = a(0)2 + b(0) + c, or c = 2.
EQN 2: 4 = a(2)2 + b(2) + c, or 4 = 4a + 2b + 2, or finally 4a + 2b = 2.
Since the derivative is continuous, and from inspection of the line segment, the slope of the line is -2, then p'(0) must equal -2. This gives us:
EQN 3: -2 = 2a(0) + b, or b = -2.
Substituting into EQN 2, we have 4a + 2(-2) = 2, or 4a = 6, or a = 3/2.
Thus, the equation is:
p(x) = (3/2)x2 - 2x + 2 |
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