Find a polynomial function whose graph passes through (6,12), (11, - 12), and (0,5).

Guest Feb 4, 2021

#1**+1 **

There are three given points for this particular polynomial. Therefore, the minimum degree of this mystery polynomial must be of degree 2. The polynomial could be of a higher degree, but there is no need to consider such an option.

Since the polynomial is of degree 2, this polynomial is a quadratic. Quadratics are of the form \(f(x) = ax^2+bx+c\). Let's use the information given to use to determine the equation. Normally, this would be a nasty three-variable system of linear equations, but we can avoid this if we recognize that the y-intercept is already given.

Since the y-intercept is already given at the point (0, 5), \(c=5\). Unfortunately, we cannot take any more shortcuts beyond this point.

\(f(6) = a*6^2 + b*6 + c\\ 36a + 6b + 5 = 12\\ \fbox{1}\; 36a + 6b = 7\\ f(11) = a*11^2 + b*11 + c\\ 121a + 11b + 5 = -12\\ \fbox{2}\;121a + 11b=-17\)

Now, let's solve this system of two variables. Unfortunately, it is already clear that this system will be an algebraic nightmare, but we will manage and perservere. I will use the elimination method and eliminate b first as that is probably the simplest one to do at this point.

\(\quad 6*\fbox{2}\quad\quad 726a+66b=-102\\ -11*\fbox{1}\; -396a-66b=-\;\;77\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\ \quad\quad\quad\quad\quad\quad 330a \quad\quad\; =-179\\ a=-\frac{179}{330}\)

Ok, let's now solve for b. Neither equation looks as if there will be an easy substitution. I chose equation 1, but either one will do.

\(\fbox{1}\;36a+6b=7\\ 36*-\frac{179}{330}+6b=7\\ -6444+1980b=2310\\ 1980b=8754\\ b=\frac{8754}{1980}=\frac{1459}{330}\)

Therefore, the equation of this polynomial is as follows: \(f(x) = -\frac{179}{330}x^2+\frac{1459}{330}x+5\). Yikes! This one is awful to do by hand.

Guest Feb 4, 2021

#1**+1 **

Best Answer

There are three given points for this particular polynomial. Therefore, the minimum degree of this mystery polynomial must be of degree 2. The polynomial could be of a higher degree, but there is no need to consider such an option.

Since the polynomial is of degree 2, this polynomial is a quadratic. Quadratics are of the form \(f(x) = ax^2+bx+c\). Let's use the information given to use to determine the equation. Normally, this would be a nasty three-variable system of linear equations, but we can avoid this if we recognize that the y-intercept is already given.

Since the y-intercept is already given at the point (0, 5), \(c=5\). Unfortunately, we cannot take any more shortcuts beyond this point.

\(f(6) = a*6^2 + b*6 + c\\ 36a + 6b + 5 = 12\\ \fbox{1}\; 36a + 6b = 7\\ f(11) = a*11^2 + b*11 + c\\ 121a + 11b + 5 = -12\\ \fbox{2}\;121a + 11b=-17\)

Now, let's solve this system of two variables. Unfortunately, it is already clear that this system will be an algebraic nightmare, but we will manage and perservere. I will use the elimination method and eliminate b first as that is probably the simplest one to do at this point.

\(\quad 6*\fbox{2}\quad\quad 726a+66b=-102\\ -11*\fbox{1}\; -396a-66b=-\;\;77\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\ \quad\quad\quad\quad\quad\quad 330a \quad\quad\; =-179\\ a=-\frac{179}{330}\)

Ok, let's now solve for b. Neither equation looks as if there will be an easy substitution. I chose equation 1, but either one will do.

\(\fbox{1}\;36a+6b=7\\ 36*-\frac{179}{330}+6b=7\\ -6444+1980b=2310\\ 1980b=8754\\ b=\frac{8754}{1980}=\frac{1459}{330}\)

Therefore, the equation of this polynomial is as follows: \(f(x) = -\frac{179}{330}x^2+\frac{1459}{330}x+5\). Yikes! This one is awful to do by hand.

Guest Feb 4, 2021