Finding the equation for the parabola

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SUMMARY

The discussion focuses on finding the coefficients of a parabola in the form y=ax^2+bx, given a tangent line y=5x-8 at the point P=(2,2). The slope of the tangent line is determined to be 5, leading to the derivative of the parabola, 2ax+b, which is set equal to 5. This results in the equation 5=4a+b. Additionally, substituting the point P into the parabola equation provides another equation to solve for the coefficients a and b.

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Homework Statement


Find the parabola of the form y=ax^2+bx, whose tangent at given point P has equation y.
y=5x-8
P=(2,2)


Homework Equations


I guess the equations involved would be the equation for the parabola: y=ax^2+bx, and the y-y1=m(x-x1) for the slope.


The Attempt at a Solution


Using the parabola equation, I found a general derivative and then input the x and y' values in order to try and find a and b, but I was unable to solve for either.
 
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Why in the world do you tell us that you did all those things without showing us exactly what you got when you did it? We can't tell where you went wrong if you don't show us what you did!

You know, I hope, that the line y= 5x- 8 has slope 5 and if x= 2, y= 2.
 
Okay, well in order to find the slope, I took the derivative of the parabola equation and got 2ax+b for the derivative. I sent it equal to 5 since that is the slope and put in 2 for x since it is given getting 5=4a+b. This is where I got lost. I didn't know what to do once I got to this point.
 
You also know that when x = 2, y = 2. This gives you another equation.
 

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