#### Answer

The slope of the curve at $x=1$ is $-1$.

#### Work Step by Step

The slope $m$ of the tangent line of the curve $f(x)$ at $A(a,b)$ is also the slope of the curve at that point, which is calculated by $$m=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$$
$$f(x)=5x-3x^2\hspace{1cm}x=1$$
To calculate the slope of the curve at $x=1$, we apply the above formula:
$$m=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$$
Here $a=x=1$
$$m=\lim_{h\to0}\frac{f(h+1)-f(1)}{h}=\lim_{h\to0}\frac{\Big(5(h+1)-3(h+1)^2\Big)-(5\times1-3\times1^2)}{h}$$
$$m=\lim_{h\to0}\frac{\Big(5h+5-3(h^2+2h+1)\Big)-2}{h}$$
$$m=\lim_{h\to0}\frac{5h+5-3h^2-6h-3-2}{h}=\lim_{h\to0}\frac{-3h^2-h}{h}$$
$$m=\lim_{h\to0}(-3h-1)$$
$$m=-3\times0-1=-1$$