Answer
$7$
Work Step by Step
Consider the tangent line that contains the point $(x, c)$. The slope the tangent line to the graph of $f(x)$ at $(x, c)$ can be written as
$f'(x) =\lim\limits_{x \to c} \dfrac{f(x)-f(c)}{x-c}$
Consider $c=1$ and $f(x)=2x^2+3x$. Thus, we have:
$f'(1) =\lim\limits_{x \to 1} \dfrac{f(x)-f(1)}{x-1} \\=\lim\limits_{x \to 1} \dfrac{2x^2+3x-5}{x-1} \\=\lim\limits_{x \to 1} \dfrac{(x-1)(2x+5)}{x-1}\\=\lim\limits_{x \to 1} 2x+5\\=7 $