Use Taylor series to find infinitely many parabolas, third-degree polynomials,

Use Taylor series to find infinitely many parabolas, third-degree polynomials, etc that all have f(x) = 3x + 5 as one of their tangent lines.
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eninsala06
Step 1
Given that, f(x)=3x+5.
Formula used:
General form of Taylor expansion centered at a $f\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{{f}^{n}\left(a\right)}{n!}{\left(x-a\right)}^{n}$.
Where ${f}^{n}$ is the n th derivative of f.
Third degree Taylor polynomial consisting of first 4 terms of the Taylor expansion.
Polynomials:
$f\left(a\right)+{f}^{\prime }\left(a\right)\left(x-a\right)+f{}^{″}\left(a\right)\frac{{\left(x-a\right)}^{2}}{2}+f{}^{″}\left(a\right)\frac{{\left(x-a\right)}^{3}}{6}$
Step 2
f(x)=3x+5
f'(x)=3
f''(x)=0
f'''(x)=0
Third degree Taylor polynomials is
3a+5+3(x-a)+0+0=3a+5+3x-3a
=3x+5
Thus, the polynomial is 3x+5.

Jeffery Autrey
f(x)=3x+5
general form of Taylor expansion centored at a
$f\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{{f}^{\left(n\right)}\left(a\right)}{n!}{\left(x-a\right)}^{n}$. Here ${f}^{\left(n\right)}$ is the ${n}^{th}$ derivative of
$f\left(a\right)+{f}^{\prime }\left(a\right)\left(x-a\right)+\frac{f{}^{″}\left(a\right)}{2}{\left(x-a\right)}^{2}+\frac{f{}^{‴}\left(a\right)}{6}{\left(x-a\right)}^{3}$
f(x)=3x+5
f'(x)=3
f''(x)=0
f'''(x)=0
3a+5+3(x-a)+0+0
=3a+5+3x-3a
=3x+5
Polynomial = 3x+5