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Questions 3
Answers 932


Given three points, the simplest equation that I can think of would be a quadratic. The standard form of a quadratic is written as \(ax^2+bx+c\) , where we can tweak a,b, and c so that it intersects the points listed above. 


\(f(x)=ax^2+bx+c\) Let's substitute in the individual values of x into these equations.
\(\boxed{1}\quad f(8)=a*8^2+b*8+c\\ \boxed{1}\quad 1=64a+8b+c\) Substitute in x=8 to create the first equation. Simplify completely.
\(\boxed{2}\quad f(10)=a*10^2+b*10+c\\ \boxed{2}\quad 4=100a+10b+c\) Substitute in x=10 to create the second equation. Simplify completely.
\(\boxed{3}\quad f(13)=a*13^2+b*13+c\\ \boxed{3}\quad 8=169a+13b+c\) Substitute in x=13 to create the second equation. Simplify completely.


Notice that we have now generated a three-variable system of equations. Let's solve this for all the variables. I tried my best to pick the best path possible. I have numbered all the equations so that you can refer to them easily. 


\(\boxed{1}\quad 1=64a+8b+c\\ \boxed{4}\quad -1=-64-8b-c\) Notice that all I did here was to negate all the sides. I do this so that I can eliminate the "c" values from two of the equations. 
\(\boxed{2}\quad 4=100a+10b+c\\ \boxed{4}\quad -1=-64a-8b-c\\ \overline{\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad}\\ \boxed{5}\quad 3=36a+2b\) I added equation 2 and 4 together to generate a new equation, 5, with one fewer variable. 


\(\boxed{3}\quad 8=169a+13b+c\\ \boxed{6}\quad -8=-169a-13b-c\) I have done the exact same process as above; I negated both sides of equation 3. This is for the same purpose; eliminate "c."
\(\boxed{6}\quad -8=-169a-13b-c\\ \boxed{2}\quad 4=100a+10b+c\\ \overline{\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad}\\ \boxed{7}-4=-69a-3b\) Yet again, I added equations 6 and 2 to create another equation with one fewer variable. 


We have now successfully eliminated "c" in two instances; it is now time to use these two new equations in order to isolate another variable. I will still use elimination. 


\(\boxed{5}\quad 3=36a+2b\\ \boxed{8}\quad 9=108a+6b\) In order to create equation 8, I multiplied both sides of equation 5 by 3.
\(\boxed{7}-4=-69a-3b\\ \boxed{9}-8=-138a-6b\) In order to create equation 9, I multiplied both sides of equation 7 by 2.
\(\boxed{8}\quad 9=108a+6b\\ \boxed{9}\quad -8=-138a-6b\\ \overline{\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad}\\ \boxed{10}1=-30a\) Finally! We have reduced this to one variable only!
\(a=-\frac{1}{30}\) We now have the value of "a," Let's use this information to help us fill out the rest.


Let's now substitute this back into one of the equations and find "b."

\(a=-\frac{1}{30}\\ \boxed{5}\quad 3=36a+2b\\ 3=36*-\frac{1}{30}+2b\) I have substituted in the value of "a" into equation 5. I can now solve for b. I will multiply both sides of this equation by 30 to eliminate the presence of those fractions.
\(90=-36+60b\\ 15=-6+10b\\ 21=10b\\ b=\frac{21}{10}\) Notice how I decided to divide the equation by 6 immediately to simplify it into more manageable numbers. While this is not strictly necessary, I think this is always a good strategy to use when equations get out of hand. Look at that! We have the value of "b!"


Equation 2 looks like the easiest one to substitute back into so that is the one that I will do. 


\(a=-\frac{1}{30}; b=\frac{21}{10}\\ \boxed{2}\quad 4=100a+10b+c\\ 4=100*-\frac{1}{30}+10*\frac{21}{10}+c\) You might be able to see why I preferred equation 2. The substitution appears to be the path of least resistance.
\(4=-\frac{10}{3}+21+c\\ -17=-\frac{10}{3}+c\\ 51=10-3c\\ 41=-3c\\ c=-\frac{41}{3}\) We have finally solved for the final unknown. 


The simplest equation I could come up with is \(f\left(x\right)=-\frac{1}{30}x^2+\frac{21}{10}x-\frac{41}{3}\)

Feb 15, 2019