Question

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

Upper level algebra
ANSWERED
asked 2020-11-01
Find the x-and y-intercepts of the graph of the equation algebraically.
\(\displaystyle{y}=-{3}{\left({2}{x}+{1}\right)}\)

Answers (1)

2020-11-02
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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