# Find the x-and y-intercepts of the graph of the equation algebraically. y=-3(2x+1)

Find the x-and y-intercepts of the graph of the equation algebraically.
$$\displaystyle{y}=-{3}{\left({2}{x}+{1}\right)}$$

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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.
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