Send me a graph with workings as an example

Let's choose some $x$ values around the vertex. The $x$-coordinate of the vertex is $x = -\frac{b}{2a} = -\frac{-4}{2(1)} = 2$.

$x$	$x^2$	$-4x$	$+3$	$y = x^2 - 4x + 3$
0	0	0	3	3
1	1	-4	3	0
2	4	-8	3	-1
3	9	-12	3	0
4	16	-16	3	3

The points to plot are $(0,3)$, $(1,0)$, $(2,-1)$, $(3,0)$, $(4,3)$.

\begin{tikzpicture} \begin{axis}[ axis lines=middle, xlabel=$x$, ylabel=$y$, xmin=-0.5, xmax=4.5, ymin=-1.5, ymax=3.5, xtick={0,1,2,3,4}, ytick={-1,0,1,2,3}, grid=both, grid style={line width=.1pt, draw=gray!10}, major grid style={line width=.2pt,draw=gray!50}, minor tick num=1, every axis x label/.style={at={(current axis.right of origin)},anchor=north west}, every axis y label/.style={at={(current axis.above origin)},anchor=south east} ] \addplot[domain=0:4, samples=100, blue, thick]{x^2 - 4*x + 3}; \addplot[only marks, mark=*, mark options={fill=red}] coordinates {(0,3) (1,0) (2,-1) (3,0) (4,3)}; \node[above right] at (axis cs:0,3) {$ (0,3) $}; \node[below left] at (axis cs:1,0) {$ (1,0) $}; \node[below right] at (axis cs:2,-1) {$ (2,-1) $}; \node[below right] at (axis cs:3,0) {$ (3,0) $}; \node[above left] at (axis cs:4,3) {$ (4,3) $}; \end{axis} \end{tikzpicture}

Given the function $y = x^2 - 4x + 3$.
Differentiate term by term with respect to $x$:

$$\frac{dy}{dx} = \frac{d}{dx}(x^2) - \frac{d}{dx}(4x) + \frac{d}{dx}(3)$$ $$\frac{dy}{dx} = 2x - 4 + 0$$ $$\frac{dy}{dx} = 2x - 4$$

Substitute $x=1$ into the gradient function:

$$\frac{dy}{dx} \Big|_{x=1} = 2(1) - 4$$ $$\frac{dy}{dx} \Big|_{x=1} = 2 - 4$$ $$\frac{dy}{dx} \Big|_{x=1} = -2$$

The gradient of the graph $y = x^2 - 4x + 3$ at $x=1$ is $\boxed{-2}$. This means the tangent line at $(1,0)$ has a slope of $-2$.