# Determine the real numbers A and B if the line with equation 10x+y+2=0 is tangent to the parabola y=Ax^2+Bx+1 at the point (-1,8).

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Given that the line `10x+y+2=0` is tangent to the parabola `y=Ax^2+Bx+1` at (-1,8), determine A and B:

The slope of the line is -10. (Write in slope-intercept form y=-10x-2). The slope of the line tangent to a curve at a given point is the first derivative of the curve evaluated at that point.

Taking the first derivative we get :

`y'=2Ax+B` . Knowing that at x=-1 the slope of the tangent line is -10 yields `-10=-2A+B` .

We also know that (-1,8) lies on the curve, so `y=Ax^2+Bx+1==>8=A(-1)^2+B(-1)+1==>7=A-B` .

Solving simultaneously we get:

`-2A+B=-10`

`A-B=7`

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`-A=-3 ==> A=3 ==>B=-4`

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**The parameters A and B are A=3,B=-4**

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Checking we have `y=3x^2-4x+1` as the equation of the parabola. The first derivative is `y'=6x-4` , which when evaluated at x=-1 gives -10 as required. Also note that `8=3(-1)^2-4(-1)+1` so the point (-1,8) is on the curve.