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January 30, 2015

January 30, 2015

Posted by **Kim** on Thursday, May 3, 2012 at 2:58am.

• In the case of a linear function f (x) = ax + b, this is easy.

– If a ≠ 0, then x = −b /a.

– If a = 0, then there is no solution unless b = 0, in which case any real number is a solution.

• In the case of a quadratic function f (x) = ax^2+ b x +c, this is relatively easy.

– If a ≠ 0, then x = (-b±√(b^2-4ac))/2a

– If a = 0, then you are really in the case of a line.

Write a brief description of Newton’s method that should explain the concept to someone who knows pre-calculus, but not Calculus.

- Calculus -
**Damon**, Thursday, May 3, 2012 at 7:23amDraw a tangent to the graph of the function, see where it hits the x axis.

Then draw a new tangent to the function at the x you just found. Continue until your function is close to zero.

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