# What is the minimum vertical distance between the parabolas y=x^2+1 and PS

What is the minimum vertical distance between the parabolas $$\displaystyle{y}={x}^{{2}}+{1}$$ and $$\displaystyle{y}={x}-{x}^{{2}}$$

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Sally Cresswell
$$\displaystyle{y}_{{1}}={x}^{{1}}+{1},\ {y}_{{2}}={x}-{x}^{{2}}$$
The vertical at any point x is obtained form
$$\displaystyle{D}={y}_{{1}}-{y}_{{2}}={x}^{{2}}+{1}-{\left({x}-{x}^{{2}}\right)}$$
$$\displaystyle={x}^{{2}}+{1}-{x}+{x}^{{2}}$$
$$\displaystyle={2}{x}^{{2}}-{x}+{1}$$
If you subtract the other way the distance you get is negative because of where the graphs are situated, but the problem would work out similarly.
Find where $$\displaystyle{D}'={0}$$
$$\displaystyle{D}'{\left({x}\right)}={0}={4}{x}-{1}$$
$$\displaystyle{x}={\frac{{{1}}}{{{4}}}}$$
$$\displaystyle{D}{\left({x}\right)}={2}{x}^{{2}}-{x}+{1}$$ is an upward opening parabola so $$\displaystyle{x}={\frac{{{1}}}{{{4}}}}$$ is the minimum.
Plug in to get the minimum distance
$$\displaystyle{D}{\left({\frac{{{1}}}{{{4}}}}\right)}={2}{\left({\frac{{{1}}}{{{4}}}}\right)}^{{2}}-{\frac{{{1}}}{{{4}}}}+{1}$$
$$\displaystyle={\frac{{{1}}}{{{8}}}}-{\frac{{{1}}}{{{4}}}}+{1}$$
$$\displaystyle={\frac{{{1}-{2}+{8}}}{{{8}}}}$$
$$\displaystyle={\frac{{{7}}}{{{8}}}}$$
Result: $$\displaystyle{\frac{{{7}}}{{{8}}}}$$