Verify a conjugacy

Show that $f(x)=x^2+(2\lambda-\lambda^2)/4$ is conjugate to $g(x)=\lambda x(1-x)$ via the conjugacy $\varphi(x) = -\lambda x + \lambda/2$.

To show that this, we must show that $f\circ \varphi=\varphi\circ g$.

\begin{align*}
f\circ\varphi&=f\Big(-\lambda x+\frac{\lambda}{2}\Big)\\
&=\Big(-\lambda x+\frac{\lambda}{2}\Big)^2+\frac{2\lambda-\lambda^2}{4}\\
&=\Big(\lambda x\Big)^2+\Big(\frac{\lambda}{2}\Big)^2-2\Big(\lambda x\cdot\frac{\lambda}{2}\Big)+\frac{2\lambda-\lambda^2}{4}\\
&=\frac{\lambda^2(4x^2)+\lambda^2+\lambda^2(-4x)-\lambda^2+2\lambda}{4}\\
&=\frac{\lambda^2(4x^2+1-4x-1)+2\lambda}{4}\\
&=\lambda^2(x^2-x)+\frac{\lambda}{2}.\\
\\
\varphi\circ g&=\varphi(\lambda x-\lambda x^2)\\
&=-\lambda(\lambda x-\lambda x^2)+\frac{\lambda}{2}\\
&=\lambda^2(x^2-x)+\frac{\lambda}{2}
\end{align*}

Since $f\circ \varphi=\varphi\circ g$, $f(x)$ is conjugate to $g(x)$ via the conjugacy $\varphi(x)$.