Step 1

Straight Line:

The given points are \(\displaystyle{\left({0},\ {7}\right)},\ {\left({8},\ {0}\right)}\)

The slope of the points is \(\displaystyle{\left({m}\right)}={\frac{{{y}_{{{2}}}-{y}_{{{1}}}}}{{{x}_{{{2}}}-{x}_{{{1}}}}}}\)

Here \((x_{1},\ y_{1})=(0,\ 7)\ \&\ (x_{2},\ y_{2})=(8,\ 0)\)

\(\displaystyle{m}={\frac{{-{7}}}{{{8}}}}\)

Therefore the equation be

\(\displaystyle{y}={m}{x}+{c}\)

\(\displaystyle{y}=-{\frac{{{7}}}{{{8}}}}{x}+{c}\)

it is pas sin g through the point (0, 7)

\(\displaystyle{7}={c}\)

\(\displaystyle{y}=-{\frac{{{7}}}{{{8}}}}{x}+{7}\)

\(\displaystyle{8}{y}+{7}{x}={7}\)

Step 2

\(\displaystyle{y}={x}^{{{2}}}-{9}\)

We need to find vertex of the parabola.

\(\displaystyle{y}+{9}={x}^{{{2}}}\)

The vertex of the parabola is (0, -9).

Step 3

\(\displaystyle{f{{\left({x}\right)}}}={x}^{{{2}}}-{8}{x}+{8}\)

For maximum and minimum value we need to find \(\displaystyle{f}'{\left({x}\right)}\) and then put \(\displaystyle{f}'{\left({x}\right)}={0}\)

Now we should check for which value \(\displaystyle{f}{''}{\left({x}\right)}{>}{0}\) or \(\displaystyle{<}{0}\).

\(\displaystyle{f}'{\left({x}\right)}={2}{x}-{8}\)

\(\displaystyle{f}'{\left({x}\right)}={0}\)

\(\displaystyle{2}{x}-{8}={0}\)

\(\displaystyle{x}={4}\)

\(\displaystyle{f}{''}{\left({x}\right)}={2}{>}{0}\) (so there is only minimum value possible)

\(\displaystyle{f{{\left({4}\right)}}}={4}^{{{2}}}-{8}\times{4}+{8}\)

\(\displaystyle=-{8}\)

This is the required minimum value.

Step 4

\(\displaystyle{f{{\left({x}\right)}}}=-{2}{x}^{{{2}}}+{8}{x}-{5}\)

For maximum and minimum value we need to find \(\displaystyle{f}'{\left({x}\right)}\) and then put \(\displaystyle{f}'{\left({x}\right)}={0}\)

Now we should check for which value \(\displaystyle{f}{''}{\left({x}\right)}{>}{0}\) or \(\displaystyle{<}{0}\).

\(\displaystyle{f}'{\left({x}\right)}=-{4}{x}+{8}\)

\(\displaystyle{f{{\left({x}\right)}}}={0}\)

\(\displaystyle-{4}{x}+{8}={0}\)

\(\displaystyle{x}={2}\)

\(\displaystyle{f}{''}{\left({x}\right)}=-{4}{<}{0}\) (so there is only maximum value possible)

\(\displaystyle{f{{\left({2}\right)}}}=-{2}\times{\left({2}\right)}^{{{2}}}+{8}\times{2}-{5}\)

\(\displaystyle={3}\)

This is the required maximum value.

This is the required minimum value.