Calculation:

Consider, the equation, \(\displaystyle{y}=-{3}{\left({2}{x}+{1}\right)}\),

To compute x-intercept, put \(\displaystyle{y}={0}\),

\(\displaystyle{0}=-{3}{\left({2}{x}+{1}\right)}\)

Multiply by \(\displaystyle-{\frac{{{1}}}{{{3}}}}\) on both the sides,

\(\displaystyle{\frac{{-{1}}}{{{3}}}}\cdot{0}={\frac{{-{3}}}{{-{3}}}}{\left({2}{x}+{1}\right)}\)

\(\displaystyle{0}={2}{x}+{1}\)

Subtract 1 on both sides,

\(\displaystyle{0}-{1}={2}{x}+{1}-{1}\)

\(\displaystyle-{1}={2}{x}\)

Divide the equation by 2,

\(\displaystyle{\frac{{-{1}}}{{{2}}}}={\frac{{{2}{x}}}{{{2}}}}\)

\(\displaystyle{x}={\frac{{-{1}}}{{{2}}}}\)

So, the x-intercept is \(\displaystyle{\left({\frac{{-{1}}}{{{2}}}},{0}\right)}\).

To compute y -intercept, put \(\displaystyle{x}={0}\),

\(\displaystyle{y}=-{3}{\left({2}{\left({0}\right)}+{1}\right)}\)

\(\displaystyle{y}=-{3}\cdot{1}\)

\(\displaystyle{y}=-{3}\)

So, the y -intercept is \(\displaystyle{\left({0},-{3}\right)}\).

Hence, x and y -intercepts of \(\displaystyle{y}=-{3}{\left({2}{x}+{1}\right)}\) are \(\displaystyle{\left({\frac{{-{1}}}{{{2}}}},{0}\right)}\) and \(\displaystyle{\left({0},-{3}\right)}\),respectively.

Consider, the equation, \(\displaystyle{y}=-{3}{\left({2}{x}+{1}\right)}\),

To compute x-intercept, put \(\displaystyle{y}={0}\),

\(\displaystyle{0}=-{3}{\left({2}{x}+{1}\right)}\)

Multiply by \(\displaystyle-{\frac{{{1}}}{{{3}}}}\) on both the sides,

\(\displaystyle{\frac{{-{1}}}{{{3}}}}\cdot{0}={\frac{{-{3}}}{{-{3}}}}{\left({2}{x}+{1}\right)}\)

\(\displaystyle{0}={2}{x}+{1}\)

Subtract 1 on both sides,

\(\displaystyle{0}-{1}={2}{x}+{1}-{1}\)

\(\displaystyle-{1}={2}{x}\)

Divide the equation by 2,

\(\displaystyle{\frac{{-{1}}}{{{2}}}}={\frac{{{2}{x}}}{{{2}}}}\)

\(\displaystyle{x}={\frac{{-{1}}}{{{2}}}}\)

So, the x-intercept is \(\displaystyle{\left({\frac{{-{1}}}{{{2}}}},{0}\right)}\).

To compute y -intercept, put \(\displaystyle{x}={0}\),

\(\displaystyle{y}=-{3}{\left({2}{\left({0}\right)}+{1}\right)}\)

\(\displaystyle{y}=-{3}\cdot{1}\)

\(\displaystyle{y}=-{3}\)

So, the y -intercept is \(\displaystyle{\left({0},-{3}\right)}\).

Hence, x and y -intercepts of \(\displaystyle{y}=-{3}{\left({2}{x}+{1}\right)}\) are \(\displaystyle{\left({\frac{{-{1}}}{{{2}}}},{0}\right)}\) and \(\displaystyle{\left({0},-{3}\right)}\),respectively.