 a) Find the equation of the line that passes through the given points. (Use x as sibuzwaW 2021-11-06 Answered
a) Find the equation of the line that passes through the given points. (Use x as your variable.)
$$\displaystyle{\left({0},\ {7}\right)},\ {\left({8},\ {0}\right)}$$
$$\displaystyle{y}=?$$
b) Find the vertex of the graph of the equation.
$$\displaystyle{y}={x}^{{{2}}}-{9}$$
$$\displaystyle{\left({x},\ {y}\right)}={()}$$
c) Find the minimum or maximum value of the quadratic function.
$$\displaystyle{f{{\left({x}\right)}}}={x}^{{{2}}}-{8}{x}+{8}$$
$$\displaystyle{()}?$$
d) Find the minimum or maximum value of the quadratic function.
$$\displaystyle{f{{\left({x}\right)}}}=-{2}{x}^{{{2}}}+{8}{x}-{5}$$
$$\displaystyle{()}?$$

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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.