Find a second-degree polynomial P such that P(2) = 2, P'(2) = 4, and P''(2) = 4 ?

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Answer 1 · 2017-09-22T23:02:27

#P(x) = 2x^2-4x+2#

First find a quadratic #f(x)# such that:

#{ (f(0) = 2), (f'(0) = 4), (f''(0) = 4) :}#

Suppose:

#f(x) = ax^2+bx+c#

Then:

#2 = f(0) = a(color(blue)(0))^2+b(color(blue)(0))+c = 0+0+c = c#

#4 = f'(0) = 2a(color(blue)(0))+b#

#4 = f''(0) = 2a#

So:

#f(x) = 2x^2+4x+2#

Define:

#P(x) = f(x-2) = 2(x-2)^2+4(x-2)+2 = 2x^2-4x+2#

Answer 2 · 2017-09-22T23:06:56

# P(x) = 2x^2 -4x + 2 #

Suppose the polynomial is:

# P(x) = Ax^2 + Bx + C #

Differentiating wrt #x# (twice) we get:

# \ \ P'(x) = 2Ax + B #
# P''(x) = 2A #

We require that:

# P''(2 ) = 4 => 2A = 4 #
# :. A = 2 #

And:

# P'(2) = 4 => 4A + B = 4 #
# :. 8+B = 4 => B = -4 #

And:

# P(2) = 2 => 4A + 2B + C =2 #
# :. 8 - 8 + C =2 => C = 2 #

Leading to:

# P(x) = 2x^2 -4x + 2 #

Validation:

# \ \ \ \P(x) = 2x^2 -4x + 2 #
# \ \ P'(x) = 4x -4 #
# P''(x) = 4 #

Then:

# \ \ \ \P(2) = 2x^2 -12x + 18 = 2#
# \ \ P'(2) = 4x -4 = 4#
# P''(2) = 4 #