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Y=ax²+bx+c that has slope 7 x=1 and slope 1 x=-2 through the point (0,-2)

User Geddski
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Alright, we'll need to find the values of a, b, and c to create our parabolic function in the form of y = ax² + bx + c. Here's how you can solve it:

First, a brief intro. The equation of a parabola is given by y = ax² + bx + c.

1. To find the slope of the parabola at any point x, we differentiate it with respect to x. The derivative with respect to x gives the slope of the function at any given point. The derivative of y (which we will denote by y') is given by y' = 2ax + b.

2. We need the parabola to pass through the point (0,-2). From that, we make use of the equation of the parabola and substitute the value of the point to y = ax² + bx + c, which becomes -2 = a(0)² + b(0) + c; simplifies to c = -2.

3. Now, given that the slope is 7 at x = 1. We substitute 1 into the derivative of the function: 7 = 2a(1) + b; simplifies to 7 = 2a + b.

4. Similarly, the slope is 1 at x = -2. We substitute -2 into the derivative: 1 = 2a(-2) + b; simplifies to 1 = -4a + b.

5. Now, we have a system of the two equations: 2a + b = 7 and -4a + b = 1. We solve this system of equations by substitution or elimination:

Subtracting these two equations eliminates b and let us find a:

2a - (-4a) = 7 - 1,

6a = 6,

a = 6 / 6 = 1.

6. Substituting a = 1 in the first equation, we find:

2 * 1 + b = 7,

2 + b = 7,

b = 7 - 2 = 5.

Finally, we have the coefficients a, b, and c defined as a=1, b=5, c=-2 respectively.

Hence, the resulting parabola that satisfies these conditions is y = x² + 5x - 2.

User Jengar
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