How to change Hermite function to have a same shape with moving extremum point and zeros

If somebody has a experience with polynomials.

How to set this Hermite function to have a general minimum where I want on $x$ axis, for example in $0.5$. Is it possible to be in analytic form $f[x,a]$? Point $a$ is minimum.

$$f[x] = -(1/48) + (3 x^2)/16 - (5 x^4)/16 + (7 x^6)/48$$

And the same thing with this Hermite function to have one of its roots where I want also in form $g[x,b]$? Point $b$ is one of zeros.

$$g[x] = -(x/2) + x^3/2$$

For Example I want to do this but without connecting two functions.

f[x] function is

 Plot[1/8 - x^2/4 + x^4/8, {x, -1, 1}]


and I moved the maximum in point a=47/133

Plot[Piecewise[{{2423393/2916000 + (831383 x)/972000 - (760627 x^2)/
972000 - (2352637 x^3)/2916000,
x < 47/133}, {-(17689/79507) + (2494149 x)/318028 - (2388015 x^2)/
159014 + (2352637 x^3)/318028, x > 47/133}}], {x, -1, 1}]


Function g[x] is

Plot[-(x/2) + x^3/2, {x, -1, 1}]


and I moved zero in point b=47/133

Plot[Piecewise[{{47/180 - (43 x)/90 - (133 x^2)/180,
x < 47/133}, {47/86 - (90 x)/43 + (133 x^2)/86,
x > 47/133}}], {x, -1, 1}]


How to do this without connecting two functions?

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Your $f$ has a global minimum in $0$. Hence, if you want to build a function $f_1$ "with the same shape" of $f$ which attains its global minimum in $a$, then you should consider the function: $$f_1(x)=f(x-a)\; .$$
The same argument shows that: $$g_1(x)=g(x-b)$$ has one root in $b$, because $g$ has one root in $0$.
In general, you can reason as follows. Assume you want to find a function $f_1$ such that 1) its graph has "the same shape" of the graph of another known function $f$ and 2) it takes the value $f(x_0)$ in the point $x_1\neq x_0$ (this value could be a max, a min, a zero, or any arbitrary number in the range of $f$), i.e. $f_1(x_1)=f(x_0)$; then the function: $$f_1(x)=f(x-(x_1-x_0))$$ is the one you are looking for.