Equation for exponential function that is tangential to y=x at x=0? Ie. For $y=(x+d)^c+f$ at any given value of $c$?

I am looking for what I believe should be a simple function. I would like a line that is tangential to $y=x$ at $x=0$ and has an exponential control.

So for example $y=x^c$ but shifted so that at $x=0$ it is tangential to $y=x$ (derivative t $x=0$ is 1).

I understand probably I need to work from:

$$y=(x+d)^c+f$$

where $d$ & $f$ are automatically calculated to shift the exponential function around and $c$ controls the exponential nature.

I would then need to automatically/programmatically solve for values of $d$ and $f$ for a given value of $c$ (my exponential control) such that the derivative of this final function (slope) is $1$ at $x=0$.

I believe I understand the correct principle but I'm not able to figure out how to get it done. Is there any help anyone can provide? I appreciate it. Thanks.

Since you want tangency, the two functions must be equal at $x = 0$,
$$ 0 = (0+d)^c + f = d^c + f \text{,} \tag{1}$$ and have the same slopes/derivatives at $x = 0$, $$ 1 = c(0+d)^{c-1} = cd^{c-1} \text{.} \tag{2}$$

If $c < 0$, $(x+d)^c+f$ is strictly monotonically decreasing (where it is defined), so never has slope $1$ and tangency is impossible.

If $c = 0$, $(2)$ is unsatisfiable. (Equivalently, a constant function never has slope $1$, so tangency is impossible.)

If $c = 1$, then $(1)$ and $(2)$ are $d+f = 0$ and $1 = 1$, so there are infinitely many choices of $d$ and $f$ that work. Of course, you're translating $y = x$ to match $y = x$ which can be done infinitely many ways.

Henceforth, we suppose $c > 0$ and $c \neq 1$.

From $(2)$, $d = (1/c)^{\frac{1}{c-1}} = c^{\frac{1}{1-c}}$. Then, from $(1)$, $f = - c^{\frac{c}{1-c}}$.