Question

# The point (x,y) lies on the graph of y=x³. Write

Functions
The point (x,y) lies on the graph of $$\displaystyle{y}={x}³$$. Write the distance d between (1, 0) and (x,y) in terms of x. Write the distance d between (0, 1) and (x,y) in terms of x.

## Expert Answers (1)

2021-08-12
Use the Distance Formula: $$\displaystyle{d}=\sqrt{{{\left({x}{2}-{x}{1}\right)}^{{2}}+{\left({y}{2}-{y}{1}\right)}^{{2}}}}$$
The distance between (1,0) and (x,y) is: $$\displaystyle{d}{1}=\sqrt{{{\left({x}-{1}\right)}^{{2}}+{\left({y}-{0}\right)}^{{2}}}}$$
$$\displaystyle{d}{1}=\sqrt{{{\left({x}-{1}\right)}^{{2}}+{y}^{{2}}}}$$
Since $$\displaystyle{y}={x}^{{3}}$$, we substitute:
$$\displaystyle{d}{1}=\sqrt{{{\left({x}^{{2}}-{2}{x}+{1}\right)}+{\left({x}^{{3}}\right)}^{{2}}}}$$
$$\displaystyle{d}{1}=\sqrt{{{\left({x}^{{2}}-{2}{x}+{1}\right)}+{x}^{{6}}}}$$
$$\displaystyle{d}{1}=\sqrt{{{\left({x}^{{6}}\right)}+{x}^{{2}}-{2}{x}+{1}}}$$
Similarly, the distance between (0,1) and (x,y) is: $$\displaystyle{d}{2}=\sqrt{{{\left({x}-{0}\right)}^{{2}}+{\left({y}-{1}\right)}^{{2}}}}$$
$$\displaystyle{d}{2}=\sqrt{{{\left({x}^{{2}}\right)}+{\left({y}-{1}\right)}^{{2}}}}$$
Since $$\displaystyle{y}={x}^{{3}}$$, we substitute: $$\displaystyle{d}{2}=\sqrt{{{\left({x}^{{2}}+{\left({x}^{{3}}-{1}\right)}^{{2}}\right)}}}$$
$$\displaystyle{d}{2}=\sqrt{{{x}^{{2}}+{\left({x}^{{6}}-{2}{x}^{{3}}+{1}\right)}}}$$
$$\displaystyle{d}{2}=\sqrt{{{x}^{{6}}-{2}{x}^{{3}}+{x}^{{2}}+{1}}}$$