Find the parabola with equation y = ax^2 + bx whose tangent line at (2, 4) has equation y = 8x − 12.

We want to get a quadratic equation by knowing its tangent line at a point. Remember that the tangent line is given by the derivative of the equation.

We will find that the parabola is:

y = 3*x^2  - 4*x

We start with the tangent line:

y₂ = 8*x - 12

The slope of this tangent line is equal to the derivative of our parabola evaluated in x = 2.

Our parabola is y = a*x^2 + b*x

Derivating we get:

y' = 2*a*x + b

Then we have:

2*a*2 + b = 8

And we also know that our parabola passes through the point (2, 4), then we must have:

4 = a*2^2 + b*2

Then we have a system of equations:

2*a*2 + b = 8

4 = a*2^2 + b*2

First we can simplify the equations:

4*a + b = 8

4 = 4*a + 2*b

We can isolate 4*a in the first equation to get:

4*a = 8 - b

Now we can replace it in the second equation:

4 = (8 - b) +  2*b

4 = 8 + b

4 - 8 = b

-4 = b

Now that we know the value of b, we can use

4*a  = 8 - b

4*a = 8 + 4 = 12

a = 12/4 = 3

Then the equation of the parabola is:

y = 3*x^2  - 4*x

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step-by-step explanation: