#2**+2 **

If we assume that "the graph consists of three line segments," then we can generate the equation of the lines.

Of course, I could use \(y=mx+b\), but that would require me to know the y-intercept of the first red line, and, unfortunately, the y-intercept is not clear-cut in the image.

I will use point-slope form instead; this form requires me to know the coordinates of **any ** two points on that line. \(y-y_1=m(x-x_1)\) is the form of point-slope form. \((x_1,y_1)\) is the coordinate of any one point of the line. In the image, I can pinpoint that \((-4,4)\text{ and }(-1,-1)\) both lie on the line I care about for this problem. Now that they are identified, I can now find the slope.

\(m=\frac{-1-4}{-1-(-4)}=\frac{-5}{3}\)

I will substitute in the point \((-1,-1)\) as my point.

Since we are trying to find \(f(-2)\) , x=-2.

\(y_1=-1;m=-\frac{5}{3};x=-2;x_1=-1\\ y-y_1=m(x-x_1)\\ y-(-1)=-\frac{5}{3}(-2-(-1))\) | It is time to solve for y! |

\(y+1=-\frac{5}{3}*-1\\ y+\frac{3}{3}=\frac{5}{3}\\ y=\frac{2}{3}\) | |

\(f(-2)=\frac{2}{3}\). You will see that this answer is consistent with the initially given diagram.

** **

TheXSquaredFactorFeb 16, 2019

#1**+1 **

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 \(f\left(x\right)=-\frac{1}{30}x^2+\frac{21}{10}x-\frac{41}{3}\)

.TheXSquaredFactorFeb 16, 2019