Math Problem Solving: Linear Functions and Quadratic Equations

1. Math Mr. Beland Period 6 Sahil Kaler

2. Let’s Start with a few Jokes. 

3. Question 1 If and , then determine: • f(5) • g(a + 4) • 3f(7) • 2f(3a + 1) – g(a-1) Let’s attempt to do –i)

4. Answer 1 (i) • f(5) 1. f(x) = 3x – 8 2. f(5) = 3(5) – 8 3. f(5) = 15 – 8 4. f(5) = 7 When we see f(x) we say it as f at x. Since the question said f(5) we replace or substitute all the Xs in the equation with a 5. Then we go ahead and solve the problem. It is important for all of us to use BEDMAS.

5. Question 1 If and , then determine: • f(5) • g(a + 4) • 3f(7) • 2f(3a + 1) – g(a-1) Let’s attempt to do – ii)

6. Answer 1 (ii) ii) g(a+4) 1. g(x)=2(x+4)2–3 2. g(a+4)=2[(a+4+4]2–3 3. g(a+4)=2(a+8)2–3 4. g(a+4)=2(a2+16a+64)-3 5. g(a+4)=2a2+32a+125 This equation is no different. Instead of a number, we replace the Xs with (a+4). First, you substitute the Xs and then solve, but remember BEDMAS. In step 2, I collected like terms which were the two 4’s and added them. Then, using my math skills I expanded the bracket of (a+8)2 and solved the question.

7. Question 1 If and , then determine: • f(5) • g(a + 4) • 3f(7) • 2f(3a + 1) – g(a-1) Let’s attempt to do – iii)

8. Answer 1 (iii) iii)3f(7) 1. f(x)=3x–8 2. f(7)=3(7)–8 3. f(7)=21-8 4. f(7)=13 5. 3f(7)=3 x 13 6. 3f(3)=39 If you look at this equation, we are substituting the Xs for 7. But some of you might be wondering what is the 3 there for. If you look carefully, the 3 is outside the bracket, which means that we will multiply our answer by 3 after we have solved f(7). The reason why we do that is because since f(7) is being multiplied by 3, we then have to do the same on the other side.

9. Question 1 If and , then determine: • f(5) • g(a + 4) • 3f(7) • 2f(3a + 1) – g(a-1) Let’s attempt to do – iv)

10. Answer 1(iv) In order to do this problem, you have to break it into 3 different components. First, you have to solve for 2f(3a+1), second, you have to solve for g(a–1). Last but not least, you have to then put the two together and solve the operation that you are told. In this case, we were told to subtract, therefore we subtracted. iv) 2f(3a+1)–g(a–1) 1. f(x) = 3x–8 2. f(3a+1)=3(3a+1)–8 3. f(3a+1)=9a–5 4. 2f(3a+1)=18a–10 1. g(x)=2(x+4)2–3 2. g(a–1)=2(a+3)2–3 3. g(a–1)=2a2+12a+15 2f(3a+1)–g(a–1) =(18a–10)–(2a2+12a+15) =–2a2+6a–25 The answer is -2a2+6a–25

11. Question 2 If f(x) = x2 – 3x – 12, then determine x such that f(x) = 16. Let’s take a look at this problem.

12. Answer 2 Don’t be scared if you got two answers, it’s okay. It’s a parabola, you will get two answers. f(x) = 16 1. 16 = x2 – 3x – 12 2. 0 = x2 – 3x – 28 3.Use Quadratic Formula x = (– b ± √b2 – 4ac) ÷ 2a 4. When you do Quadratic Formula, you will get two answers, x = 7 and x = – 4. Therefore, when f(x) = 16, x will either be 7 or -4.

13. Question 3 A teacher gives a test to her students. The highest mark in the class is 180 and the lowest mark in the class is 60. Determine a linear function such the highest mark in the class is 95 and the lowest mark in the class is 55. Let’s take a look at this problem. 

14. Answer 3 This is a word problem, therefore, we have to understand the question first. Basically, the question is saying that if we input 180, the output is 95, and if we input 60, the output is 55. So, now we have to come up with a function. First we know that the two points we have are (180,95) and (60,55). Since we are told that this function is a linear function, we will try to solve this using the linear equation y = mx + b.

15. Answer 3 Let “x”be the mark got in class. Let’s find slope first. m = (y2 – y1 )/(x2 – x1) m = (95 – 55)/(180 – 60) m = 1/3 f(x) = 1/3x + b Let’s find b. We subst. a point (60,55) 55 = 1/3(60) + b b = 55 – 20 b = 35 Therefore, the function is: f(x) = 1/3x + 35

16. Thank You! Thank You for viewing my presentation! Hope you all have a great day!