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Find a function y=ax^2+bx+c whose graph has xint = 1, yint = -2 and a tangent line with slope = -1 at the yint?
2 Answers
- Anonymous1 decade agoFavorite Answer
y = ax^2 + bx + c
y' = 2ax + b
The x-intercept gives us the point (1, 0), the y-intercept gives us the point (0, -2). So we have:
0 = a(1)^2 + b(1) + c
==> a + b + c = 0
-2 = a(0) + b(0) + c
==> c = -2
Since the slope is -1 at the y-intercept, y' = 1 at x = 0.
1 = 2a(0) + b
==> b = 1
So:
a + b + c = 0
==> a + 1 - 2 = 0
==> a = 1
Therefore, the function is:
y = x^2 + x - 2
I hope this helps!
- 5 years ago
b should be -1, because they gave you the slope of the tangent line, not the original function