How do you find a parabola with equation #y=ax^2+bx+c# that has slope 4 at x=1, slope 8 at x=1 and passes through (2,15)?
The equation is
and
Adding the 2 equations, we get
Therefore,
So,
Finally
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#y=3x^22x+7#
Given 
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To find the equation of the parabola, you need to solve for the coefficients (a), (b), and (c) in the equation (y = ax^2 + bx + c).
Given that the slope of the parabola at (x = 1) is (4), you can differentiate the equation with respect to (x) and set (x = 1). Similarly, for (x = 1), the slope is (8). You can set up two equations using these conditions.

Differentiate (y = ax^2 + bx + c) with respect to (x): [y' = 2ax + b]

Substitute (x = 1) and (y' = 4): [4 = 2a(1) + b]

Substitute (x = 1) and (y' = 8): [8 = 2a(1) + b]

Solve the system of equations to find (a) and (b).

Once you have (a) and (b), substitute (x = 2) and (y = 15) into the equation (y = ax^2 + bx + c) to find (c).

Write the equation of the parabola as (y = ax^2 + bx + c) with the values of (a), (b), and (c) you found.
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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