What are the x and y intercepts of $- 3 y = 2 x^{3} - 3$ $-3y=2x^3-3$ ?

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Answer 1 · 2016-03-07T09:06:42

Intercept on $x$ $x$ axis is $1.1447$ $1.1447$ and intercept on $y$ $y$ axis is $1$ $1$ .

To find $x$ $x$ intercepts of $- 3 y = 2 x^{3} - 3$ $−3y=2x^3−3$ , one needs to put $y = 0$ $y=0$ in the equation which gives us

$- 3 \times 0 = 2 x^{3} - 3$ $−3xx0=2x^3−3$ or $2 x^{3} - 3 = 0$ $2x^3-3=0$ or $x = \frac{\sqrt[3]{3}}{2} = 1.1447$ $x=root(3)3/2=1.1447$ .

For $y$ $y$ intercepts, put $x = 0$ $x=0$ , i.e. $- 3 y = 0 - 3 = - 3$ $-3y=0-3=-3$ or $y = 1$ $y=1$

Hence, intercept on $x$ $x$ axis is $1.1447$ $1.1447$ and intercept on $y$ $y$ axis is $1$ $1$ .

What are the x and y intercepts of -3y=2x^3-3−3y=2x3−3-3y=2x^3-3?