For the mean value theorem, we know it has to be continuous, so we're going to let X = 0 also be equivalent to when 0 less than X less than two. So the four has to equal negative X ^2 + 5 X plus a when X is 0. Sticking zero in for X, we get a equal 4. We also know that at 2 the negative X ^2 + 5 X plus a has to equal MX plus B. So if we stick into we get -2 negative the quantity 2 ^2 + 5 * 2 plus the a We found a moment ago, A4 equaling m * 2 + b or we get 10 equal to m + b, two variables. We can't solve that without a second equation. So now we're going to go to the fact that the derivative F prime of X has to be continuous, also defined. So zero at X = 0, negative two, X + 5 from 0 less than X less than two, and M at 2 less than or equal to X less than or equal to 3:00. So at the two location, negative two X + 5 is going to equal M at and sticking in two we get -2 * 2. Negative 4 + 5 is going to give us one. So if M is one. Now going back to the other equation, we can stick in M is one and figure out our B being 8.