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Calculus help (finding an equation of a parabola)?
Find an equation of the parabola ax^2 + bx + c that passes through (0,1) and is tangent to the line y=x-1 at (1,0). I have no clue where to start.
Please and thank you very much.
1 Answer
- MechEng2030Lv 78 years agoFavorite Answer
Hi!
At (1, 0), the slope of the line y = x - 1 is 1, so you know that the slope of the tangent at x = 1 must be 1 (dy/dx = 1). You also know that the curve passes through (0, 1), so y = 1 when x = 0. A similar approach to (1, 0). Let's see what we get:
Plugging in (0, 1) into the function => c = 1
dy/dx = 2ax + b
Plugging in x = 1 and equating it to 1:
2a + b = 1
Plugging in (1, 0) into the function =>
a + b + c = 0 => a + b = -1
Solving, we get:
a = 2, b = -3
So the function is y = 2x^2 - 3x + 1
