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Finding the equation of a parabola? (Calculus)?
Find an equation of the parabola y = ax^2 + bx + c that passes through (0,1) and is tangent to the line y = x - 1 at (1,0)
I don't even know where to start :(...we didnt even do this in class
1 Answer
- DWReadLv 710 years ago
y = ax² + bx + c passes through (0, 1) and (1, 0).
1 = a·0² + b·0 + c
c = 1
0 = a·1² + b·1 + 1
a = -b - 1
Parabola and line have same slope at the point of tangency.
slope of line = 1
slope of parabola = y' = 2ax + b
point of tangency: (1, 0)
2a·1 + b = 1
2(-b - 1) + b = 1
-b = 3
b = -3
a = 2
y = 2x² - 3x + 1
