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

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

Demi-Leigh Barrera
Answered 2021-11-07 Author has 24846 answers

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.

Not exactly what you’re looking for?
Ask My Question
0
 

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more
...