Does there exists a quadratic function for every parabola?

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ritwik06
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Homework Statement



While going through a question, I came across a function [tex]f(x)=5^{x}+5^{-x}[/tex]
I saw its graph through a software. It was a parabola with minimum value 2.

Now a question arises in my mind.
Every function of the type [tex]g(x)=ax^{2}+bx+c[/tex] is a parabola.
Can I assume the corollary to be true, that is for every parabola, there exists a quadratic function??
If yes, how may I find the coefficients a,b,c such that f(x)=g(x) ?



The Attempt at a Solution


There is only one thing that I see-
[tex]\frac{-\Delta}{4a}=2[/tex]


Can this be solved?
 
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3. The Attempt at a Solution
For f(x)
if x=0, f(x)=2
if x=1, f(x)=5.2
if x=-1, f(x)=5.2
And if I use these equations to solve for the quadratics to solve for a,b,c the coefficients of g(x), I find that a=3.2, b=0, c=2.
which makes [tex]g(x)=3.2x^{2}+2[/tex] But the graph for this does not exactly coincide with f(x). Why??
Somebody Please help me.
 
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Th graph of

[tex] 5^x + 5^{-x}[/tex]

is not exactly a parabola, so your attempt simply gives an approximation of this function and its graph, but will not duplicate it.
 


statdad said:
Th graph of

[tex] 5^x + 5^{-x}[/tex]

is not exactly a parabola, so your attempt simply gives an approximation of this function and its graph, but will not duplicate it.

What is the definition of parabola?
 


A parabola is the graph of a function that has the form

[tex] f(x) = ax^2 + bx + c[/tex]

If you graph

[tex] x = ay^2 + by + c[/tex]

you get a parabola shape, but this is not a function.

The equation you encountered (and its graph) are a form of a catenary . The classical equation for this graph involves the hyperbolic cosine ([tex]\cosh[/tex]), or exponentials base [tex]e[/tex], but the form you give works as well. A catenary can be loosely described as the shape a hanging chain takes (or the graph of power lines between towers).