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