Find a parabola with the equation y=ax^2+bx+c and has slope 4 at x=1, slope -8 at x=-1, and passes through the point (2,15).
Thanks
Hint: take the derivative of y, then use the given slopes to find a,b Then use the point given to find c.
To find the parabola with the given conditions, we need to determine the values of the coefficients a, b, and c in the equation y = ax^2 + bx + c.
First, let's find the equation of the tangent line at x = 1 using the given slope of 4. The slope of a tangent line at any given point (x, y) on a curve is equal to the derivative of the curve at that point. Therefore, we have:
dy/dx = 4
Taking the derivative of the equation y = ax^2 + bx + c with respect to x, we get:
2ax + b = 4
Plugging in the value x = 1, we have:
2a + b = 4 ----(1)
Similarly, for the tangent line at x = -1 with a slope of -8, we have:
dy/dx = -8
Differentiating the equation y = ax^2 + bx + c, we get:
2ax + b = -8
Substituting x = -1 into the equation, we have:
-2a + b = -8 ----(2)
To find the parabola equation, we also need to substitute the point (2, 15) into the equation y = ax^2 + bx + c. Plugging in x = 2 and y = 15, we have:
15 = 4a + 2b + c ----(3)
Now we have a system of three equations with three unknowns (a, b, and c). We can solve this system of equations to find the values of a, b, and c.
Solving equations (1) and (2) simultaneously, we can eliminate b by subtracting equation (2) from equation (1):
2a + b - (-2a + b) = 4 - (-8)
2a + b + 2a - b = 4 + 8
4a = 12
a = 3
Substituting the value of a into equation (1):
2(3) + b = 4
6 + b = 4
b = -2
Finally, substituting the values of a = 3 and b = -2 into equation (3), we can solve for c:
15 = 4(3) + 2(-2) + c
15 = 12 - 4 + c
15 = 8 + c
c = 15 - 8
c = 7
Therefore, the equation of the parabola that satisfies the given conditions is:
y = 3x^2 - 2x + 7