+31 So I'm writing a program, but I can't figure out this one algorithm. I'll write it like this
When x = 8, f(x) = 1
When x = 10, f(x) = 4
When x = 13, f(x) = 8
What is the SIMPLEST equation that would make all three of these points true? Thank you
+2446 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 \(f\left(x\right)=-\frac{1}{30}x^2+\frac{21}{10}x-\frac{41}{3}\)
+2446 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 \(f\left(x\right)=-\frac{1}{30}x^2+\frac{21}{10}x-\frac{41}{3}\)
