Do the following for the points $$ (-5,2),(-2,1),(-1,1),(4,1),(5,-2) $$ : (If you are entering decimal approximations, enter at least five decimal places.) (a) Find the equation for the best-fitting parabola $$ y=a x^{2}+b x+c $$ for these points: $$ y= $$ (b) Find the equation for the best-fitting parabola with no constant term $$ y=a x^{2}+b x $$ for these points: $$ y= $$ (c) Find the equation for the best-fitting parabola with no linear term $$ y=a x^{2}+c $$ for these points: $$ y= $$ Submit answer Next item

Question

UpStudy AI · Accepted Answer

Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}2=a\left(-5\right)^{2}+c\1=a\left(-2\right)^{2}+c\1=a\left(-1\right)^{2}+c\1=a\times 4^{2}+c\-2=a\times 5^{2}+c\end{array}\right.$$
- step1: Calculate:
  $$\left\{ \begin{array}{l}2=25a+c\1=4a+c\1=a+c\1=16a+c\-2=25a+c\end{array}\right.$$
- step2: Solve the equation:
  $$\left\{ \begin{array}{l}2=25a+c\1=4a+c\a=1-c\1=16a+c\-2=25a+c\end{array}\right.$$
- step3: Substitute the value of $$a:$$
  $$\left\{ \begin{array}{l}2=25\left(1-c\right)+c\1=4\left(1-c\right)+c\1=16\left(1-c\right)+c\-2=25\left(1-c\right)+c\end{array}\right.$$
- step4: Simplify:
  $$\left\{ \begin{array}{l}2=25-24c\1=4-3c\1=16-15c\-2=25-24c\end{array}\right.$$
- step5: Solve the equation:
  $$\left\{ \begin{array}{l}c=\frac{23}{24}\1=4-3c\1=16-15c\-2=25-24c\end{array}\right.$$
- step6: Substitute the value of $$c:$$
  $$\left\{ \begin{array}{l}1=4-3\times \frac{23}{24}\1=16-15\times \frac{23}{24}\-2=25-24\times \frac{23}{24}\end{array}\right.$$
- step7: Simplify:
  $$\left\{ \begin{array}{l}1=\frac{9}{8}\1=\frac{13}{8}\-2=2\end{array}\right.$$
- step8: Calculate:
  $$\left\{ \begin{array}{l}a \in \varnothing \c \in \varnothing \end{array}\right.$$
- step9: Rewrite:
  $$(a, c) \in \varnothing$$
Solve the system of equations $$ 2 = a(-5)^2 + b(-5); 1 = a(-2)^2 + b(-2); 1 = a(-1)^2 + b(-1); 1 = a(4)^2 + b(4); -2 = a(5)^2 + b(5) $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}2=a\left(-5\right)^{2}+b\left(-5\right)\1=a\left(-2\right)^{2}+b\left(-2\right)\1=a\left(-1\right)^{2}+b\left(-1\right)\1=a\times 4^{2}+b\times 4\-2=a\times 5^{2}+b\times 5\end{array}\right.$$
- step1: Calculate:
  $$\left\{ \begin{array}{l}2=25a-5b\1=4a-2b\1=a-b\1=16a+4b\-2=25a+5b\end{array}\right.$$
- step2: Solve the equation:
  $$\left\{ \begin{array}{l}2=25a-5b\1=4a-2b\a=1+b\1=16a+4b\-2=25a+5b\end{array}\right.$$
- step3: Substitute the value of $$a:$$
  $$\left\{ \begin{array}{l}2=25\left(1+b\right)-5b\1=4\left(1+b\right)-2b\1=16\left(1+b\right)+4b\-2=25\left(1+b\right)+5b\end{array}\right.$$
- step4: Simplify:
  $$\left\{ \begin{array}{l}2=25+20b\1=4+2b\1=16+20b\-2=25+30b\end{array}\right.$$
- step5: Solve the equation:
  $$\left\{ \begin{array}{l}b=-\frac{23}{20}\1=4+2b\1=16+20b\-2=25+30b\end{array}\right.$$
- step6: Substitute the value of $$b:$$
  $$\left\{ \begin{array}{l}1=4+2\left(-\frac{23}{20}\right)\1=16+20\left(-\frac{23}{20}\right)\-2=25+30\left(-\frac{23}{20}\right)\end{array}\right.$$
- step7: Simplify:
  $$\left\{ \begin{array}{l}1=\frac{17}{10}\1=-7\-2=-\frac{19}{2}\end{array}\right.$$
- step8: Calculate:
  $$\left\{ \begin{array}{l}a \in \varnothing \b \in \varnothing \end{array}\right.$$
- step9: Rewrite:
  $$(a, b) \in \varnothing$$
Solve the system of equations $$ 2 = a(-5)^2 + b(-5) + c; 1 = a(-2)^2 + b(-2) + c; 1 = a(-1)^2 + b(-1) + c; 1 = a(4)^2 + b(4) + c; -2 = a(5)^2 + b(5) + c $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}2=a\left(-5\right)^{2}+b\left(-5\right)+c\1=a\left(-2\right)^{2}+b\left(-2\right)+c\1=a\left(-1\right)^{2}+b\left(-1\right)+c\1=a\times 4^{2}+b\times 4+c\-2=a\times 5^{2}+b\times 5+c\end{array}\right.$$
- step1: Calculate:
  $$\left\{ \begin{array}{l}2=25a-5b+c\1=4a-2b+c\1=a-b+c\1=16a+4b+c\-2=25a+5b+c\end{array}\right.$$
- step2: Solve the equation:
  $$\left\{ \begin{array}{l}2=25a-5b+c\1=4a-2b+c\a=1+b-c\1=16a+4b+c\-2=25a+5b+c\end{array}\right.$$
- step3: Substitute the value of $$a:$$
  $$\left\{ \begin{array}{l}2=25\left(1+b-c\right)-5b+c\1=4\left(1+b-c\right)-2b+c\1=16\left(1+b-c\right)+4b+c\-2=25\left(1+b-c\right)+5b+c\end{array}\right.$$
- step4: Simplify:
  $$\left\{ \begin{array}{l}2=25+20b-24c\1=4+2b-3c\1=16+20b-15c\-2=25+30b-24c\end{array}\right.$$
- step5: Solve the equation:
  $$\left\{ \begin{array}{l}b=\frac{-23+24c}{20}\1=4+2b-3c\1=16+20b-15c\-2=25+30b-24c\end{array}\right.$$
- step6: Substitute the value of $$b:$$
  $$\left\{ \begin{array}{l}1=4+2\times \frac{-23+24c}{20}-3c\1=16+20\times \frac{-23+24c}{20}-15c\-2=25+30\times \frac{-23+24c}{20}-24c\end{array}\right.$$
- step7: Simplify:
  $$\left\{ \begin{array}{l}1=4+\frac{-23+24c}{10}-3c\1=-7+9c\-2=25+\frac{3\left(-23+24c\right)}{2}-24c\end{array}\right.$$
- step8: Solve the equation:
  $$\left\{ \begin{array}{l}c=\frac{7}{6}\1=-7+9c\-2=25+\frac{3\left(-23+24c\right)}{2}-24c\end{array}\right.$$
- step9: Substitute the value of $$c:$$
  $$\left\{ \begin{array}{l}1=-7+9\times \frac{7}{6}\-2=25+\frac{3\left(-23+24\times \frac{7}{6}\right)}{2}-24\times \frac{7}{6}\end{array}\right.$$
- step10: Simplify:
  $$\left\{ \begin{array}{l}1=\frac{7}{2}\-2=\frac{9}{2}\end{array}\right.$$
- step11: Calculate:
  $$\left\{ \begin{array}{l}a \in \varnothing \b \in \varnothing \c \in \varnothing \end{array}\right.$$
- step12: Rewrite:
  $$(a, b, c) \in \varnothing$$
It seems that there was an issue with finding the best-fitting parabolas for the given points. Let's go through the calculations step by step manually to find the equations for the best-fitting parabolas.

### Given Points:
1. $$ (-5, 2) $$
2. $$ (-2, 1) $$
3. $$ (-1, 1) $$
4. $$ (4, 1) $$
5. $$ (5, -2) $$

### (a) Best-Fitting Parabola $$ y = ax^2 + bx + c $$

To find the coefficients $$ a $$, $$ b $$, and $$ c $$, we can set up a system of equations based on the points provided. We will use the method of least squares to minimize the error.

The normal equations for fitting a quadratic polynomial are derived from the following:

1. $$ \sum y = n c + b \sum x + a \sum x^2 $$
2. $$ \sum xy = c \sum x + b \sum x^2 + a \sum x^3 $$
3. $$ \sum x^2y = c \sum x^2 + b \sum x^3 + a \sum x^4 $$

Calculating the necessary sums:

- $$ n = 5 $$ (number of points)
- $$ \sum x = -5 - 2 - 1 + 4 + 5 = 1 $$
- $$ \sum y = 2 + 1 + 1 + 1 - 2 = 3 $$
- $$ \sum x^2 = 25 + 4 + 1 + 16 + 25 = 71 $$
- $$ \sum xy = (-5)(2) + (-2)(1) + (-1)(1) + (4)(1) + (5)(-2) = -10 - 2 - 1 + 4 - 10 = -19 $$
- $$ \sum x^2y = 25(2) + 4(1) + 1(1) + 16(1) + 25(-2) = 50 + 4 + 1 + 16 - 50 = 21 $$
- $$ \sum x^3 = -125 - 8 - 1 + 64 + 125 = 55 $$
- $$ \sum x^4 = 625 + 16 + 1 + 256 + 625 = 1563 $$

Now we can set up the normal equations:

1. $$ 3 = 5c + 1b + 71a $$
2. $$ -19 = 1c + 55b + 5a $$
3. $$ 21 = 71c + 55b + 1563a $$

Let's solve this system of equations.

### (b) Best-Fitting Parabola $$ y = ax^2 + bx $$

For this case, we will set $$ c = 0 $$ in the previous equations:

1. $$ 3 = 0 + 1b + 71a $$
2. $$ -19 = 0 + 55b + 5a $$
3. $$ 21 = 0 + 55b + 1563a $$

### (c) Best-Fitting Parabola $$ y = ax^2 + c $$

For this case, we will set $$ b = 0 $$ in the previous equations:

1. $$ 3 = 0 + 0 + 71a $$
2. $$ -19 = 0 + 0 + 5a $$
3. $$ 21 = 0 + 0 + 1563a $$

Now, I will solve these systems of equations to find the coefficients $$ a $$, $$ b $$, and $$ c $$.
Solve the system of equations by following steps:
- step0: Solution:
  $$\left\{ \begin{array}{l}3=71a\-19=5a\21=1563a\end{array}\right.$$
- step1: Calculate:
  $$\left\{ \begin{array}{l}a=\frac{3}{71}\a=-\frac{19}{5}\a=\frac{7}{521}\end{array}\right.$$
- step2: Find the intersection:
  $$a \in \varnothing $$
Solve the system of equations $$ 3 = 1b + 71a; -19 = 55b + 5a; 21 = 55b + 1563a $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}3=1\times b+71a\-19=55b+5a\21=55b+1563a\end{array}\right.$$
- step1: Calculate:
  $$\left\{ \begin{array}{l}3=b+71a\-19=55b+5a\21=55b+1563a\end{array}\right.$$
- step2: Solve the equation:
  $$\left\{ \begin{array}{l}b=3-71a\-19=55b+5a\21=55b+1563a\end{array}\right.$$
- step3: Substitute the value of $$b:$$
  $$\left\{ \begin{array}{l}-19=55\left(3-71a\right)+5a\21=55\left(3-71a\right)+1563a\end{array}\right.$$
- step4: Simplify:
  $$\left\{ \begin{array}{l}-19=165-3900a\21=165-2342a\end{array}\right.$$
- step5: Solve the equation:
  $$\left\{ \begin{array}{l}a=\frac{46}{975}\21=165-2342a\end{array}\right.$$
- step6: Substitute the value of $$a:$$
  $$21=165-2342\times \frac{46}{975}$$
- step7: Simplify:
  $$21=\frac{53143}{975}$$
- step8: Calculate:
  $$\left\{ \begin{array}{l}a \in \varnothing \b \in \varnothing \end{array}\right.$$
- step9: Rewrite:
  $$(a, b) \in \varnothing$$
Solve the system of equations $$ 3 = 5c + 1b + 71a; -19 = 1c + 55b + 5a; 21 = 71c + 55b + 1563a $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}3=5c+1\times b+71a\-19=1\times c+55b+5a\21=71c+55b+1563a\end{array}\right.$$
- step1: Calculate:
  $$\left\{ \begin{array}{l}3=5c+b+71a\-19=c+55b+5a\21=71c+55b+1563a\end{array}\right.$$
- step2: Solve the equation:
  $$\left\{ \begin{array}{l}b=3-5c-71a\-19=c+55b+5a\21=71c+55b+1563a\end{array}\right.$$
- step3: Substitute the value of $$b:$$
  $$\left\{ \begin{array}{l}-19=c+55\left(3-5c-71a\right)+5a\21=71c+55\left(3-5c-71a\right)+1563a\end{array}\right.$$
- step4: Simplify:
  $$\left\{ \begin{array}{l}-19=-274c+165-3900a\21=-204c+165-2342a\end{array}\right.$$
- step5: Solve the equation:
  $$\left\{ \begin{array}{l}a=\frac{92-137c}{1950}\21=-204c+165-2342a\end{array}\right.$$
- step6: Substitute the value of $$a:$$
  $$21=-204c+165-2342\times \frac{92-137c}{1950}$$
- step7: Multiply the terms:
  $$21=-204c+165-\frac{1171\left(92-137c\right)}{975}$$
- step8: Swap the sides:
  $$-204c+165-\frac{1171\left(92-137c\right)}{975}=21$$
- step9: Multiply both sides of the equation by LCD:
  $$\left(-204c+165-\frac{1171\left(92-137c\right)}{975}\right)\times 975=21\times 975$$
- step10: Simplify the equation:
  $$-38473c+53143=20475$$
- step11: Move the constant to the right side:
  $$-38473c=20475-53143$$
- step12: Subtract the numbers:
  $$-38473c=-32668$$
- step13: Change the signs:
  $$38473c=32668$$
- step14: Divide both sides:
  $$\frac{38473c}{38473}=\frac{32668}{38473}$$
- step15: Divide the numbers:
  $$c=\frac{32668}{38473}$$
- step16: Substitute the value of $$c:$$
  $$a=\frac{92-137\times \frac{32668}{38473}}{1950}$$
- step17: Calculate:
  $$a=-\frac{480}{38473}$$
- step18: Substitute the values of $$c,a:$$
  $$b=3-5\times \frac{32668}{38473}-71\left(-\frac{480}{38473}\right)$$
- step19: Calculate:
  $$b=-\frac{13841}{38473}$$
- step20: Calculate:
  $$\left\{ \begin{array}{l}a=-\frac{480}{38473}\b=-\frac{13841}{38473}\c=\frac{32668}{38473}\end{array}\right.$$
- step21: Check the solution:
  $$\left\{ \begin{array}{l}a=-\frac{480}{38473}\b=-\frac{13841}{38473}\c=\frac{32668}{38473}\end{array}\right.$$
- step22: Rewrite:
  $$\left(a,b,c\right) = \left(-\frac{480}{38473},-\frac{13841}{38473},\frac{32668}{38473}\right)$$
Here are the results for the best-fitting parabolas based on the calculations:

### (a) Best-Fitting Parabola $$ y = ax^2 + bx + c $$

The coefficients are:
- $$ a \approx -0.012476 $$
- $$ b \approx -0.359759 $$
- $$ c \approx 0.849115 $$

Thus, the equation for the best-fitting parabola is:
$$
y = -0.012476 x^2 - 0.359759 x + 0.849115
$$

### (b) Best-Fitting Parabola $$ y = ax^2 + bx $$

It appears that there is no solution for this case, indicating that the data does not fit a parabola of this form.

### (c) Best-Fitting Parabola $$ y = ax^2 + c $$

Similarly, there is no solution for this case as well, indicating that the data does not fit a parabola of this form.

### Summary of Results:
- **(a)** $$ y = -0.012476 x^2 - 0.359759 x + 0.849115 $$
- **(b)** No solution
- **(c)** No solution

If you have any further questions or need additional assistance, feel free to ask!
(a) $$ y = -0.012476x^2 - 0.359759x + 0.849115 $$ (b) No solution (c) No solution

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