QUADRATIC FUNCTIONS. ( x 7)(5x 6) = 2. Exercises: 1 3x 5 Sum: 8. We ll expand it by using the distributive property; 9. Let s use the FOIL method;

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1 QUADRATIC FUNCTIONS A. Eercises: = + = + + = +. ( 1)(3 5) (3 5) 1(3 5) = = + = +. ( 7)(5 6) (5 6) 7(5 6) Sum: ( + 1)(3 5) = Sum: ( 7)(5 6) = = + + = +. ( 1)(3 5) = + = +. ( 7)(5 6) = + = + + = ( ) ( ) ( ) 8. We ll epand it by using the distributive property; (10+ 1)(10 1) = 10 (10 1) + 1(10 1) = = Let s use the FOIL method; ( a + b)( a b) = a ab + ba b = a b.

2 10. To factor + 3 4, we need to find p and q so that pq = 4 and p+ q= 3.If you try some possibilities for pq = 4, you ll see that when p = 4 and q = 1 both conditions are satisfied. Hence, = ( + 4)( 1). 11. To factor the epression , we need to find r, p, s and q so that = ( r+ p)( s+ q). That is, rs = 3, ps + rq = 4 and pq = 1. Let s try the easiest case; p= q= 1 and r = 3, s= 1. Two of the conditions are satisfied ( rs = 3 and pq = 1), but we need to check the third: ps + rq = (1)(1) + (3)(1) = 4. So, these are the numbers we re looking for. Hence, = (3+ 1)( + 1) is an epression which is suitable to use the difference of two squares. Notice that 4 = 16. Hence, 16 = ( 4)( + 4) is also suitable for the difference of two squares. Since 100 = ( 10)( + 10). 10 = 100, we get: 14. Firstly, notice that we can not use the difference of two squares here (basically since there is no difference here, this is a sum). To factor + 100, we need p and q so that pq = 100 and p+ q= 0. There are lots of couples of numbers satisfying the first condition, but it is important to remember that to have p+ q= 0, p and q must be of opposite sign (we know that they are not 0 since their product is not). However, if p and q have opposite signs then their product should be negative. For eample, p= 10, q= 10 pq= 100. Hence, it is impossible to find such two numbers. The epression coefficients is not a product of linear factors with integer 15. To factor + + 6, we need p and q so that pq = 6 and p + q = 1. p and q so that pq = 6 p + q 1 and 6 7 and 3 5 Hence, as you can observe from the table, does not have linear factors with integer coefficients.

3 16. In Eercise-9, we epanded the epression ( a + b)( a b) and we got + =. So, the factors of ( a b)( a b) a b a a b = ( a+ b)( a b). b are ( a + b) and ( a b) ; 17. We know from Eercise-10 that only if + 4= 0 or 1= 0. Hence, the solutions of the equation = ( + 4)( 1). So, + 3 4= 0 are = 4 and = = 0 if and 18. We know that = (3+ 1)( + 1) (See Eercise-11). Hence, and only if 3+ 1= 0 or + 1= 0. 1 That is, the solutions to = 0 are = and = = 0 if 19.We have 16 = ( 4)( + 4), so 16 = 0 if and only if 4= 0 or + 4= 0. Thus, the solutions of the epression 16 = 0 are = 4 and = Since 100 = ( 10)( + 10), Hence, the solutions to 100 = 0 are = 10 and = = 0 if and only if 10 = 0 or + 10 = As we saw before, a b = ( a+ b)( a b). That is, a b = 0 if and only if a + b = 0 or a b = 0. b b a + b = 0 a = b = and a b = 0 a = b =. a a b b Hence, the solutions to the equation a b = 0 are = and =. a a B. Graphs of Quadratic Functions Eercises: 1. The quadratic formula tells us that are: b± b 4ac =. For a + 4= 0, the solutions

4 1± 1 4( 4) 1± 17 = =.. The solutions of the equation + 7= 0 are: b± b 4ac 1± 1 4()( 7) 1± 57 = = =. a () 4 3. The equation = 5 can be written as quadratic formula. The solutions are: 5= 0. Now, we can use the b± b 4ac ± 4 4(1)( 5) ± 4 = = = = 1± 6. a (1) 4. = = 0. The solutions are: b± b 4ac 3 ± 9 4()( 11) 3± 97 = = =. a () 4 5. The graph of direction. = is obtained by shifting the graph of f( ) 4 y = -4 units in the y

5 We can easily see from the graph that the intercepts are = ±, the y intercept is y = 4, and the verte is (0, 4). We can also use the following reasoning to find the verte and the intercepts (which is very useful when we do not know the graph of the function): b 0 The coordinate of the verte of f( ) = 4is given by = = 0. Since a f (0) = 4, the verte is (0, 4). The intercepts are the solutions of 4= 0. By factoring or by using the quadratic formula, we can easily get that = ± are the solutions and hence the intercepts. The y intercept is f (0) = It is not easy to draw the graph of f( ) = 4+ 5by using y =. So, this time let s firstly find the verte and intercepts and then draw the graph accordingly. b 4 The coordinate of the verte of f( ) = 4+ 5is given by = = 1. Since a 4 f (1) = (1) 4(1) + 5 = 3, the verte is (1,3). The intercepts are the solutions of 4+ 5= 0. However, for this equation the discriminant is b 4ac= 16 4()(5) = 4< 0 which means that there are no solutions. Hence, there are no intercepts. Since there are no intercepts, we need another point on the graph to be able to draw it. We have f (0) = 5, so the y intercept is (0,5). Now, since a > 0, draw a parabola with verte (1,3) that turns up and passes through (0,5). Remember that the arms of the parabola are symmetric.

6 = = + + = +. Hence, the graph of f( ) 3 ( 3) ( 1) = = + can be obtained by shifting the graph of f( ) 3 ( 1) f ( ) = firstly 1 units in the direction and then units in the y direction. The verte is : ( 1, ) and the parabola is turning down. And this shows that there are no intercepts. The y intercept is y = f(0) =

7 8. To draw the graph of f( ) = 3 + 1: Scale f ( ) = by 3 to get f ( ) = 3 Shift f ( ) = 3 1 units in y direction As it is seen from the graph, the verte is (0,1) and the intercepts are =±. The y intercept is y = 1. C. Applications 1. Their fied costs are $10,000 per month and each pair costs $0, so the monthly cost is: C ( ) = 10,

8 They can sell pairs of sandals each month at a price of 40, then their monthly 100 revenue is: R ( ) = 30 = Thus, the profit function is: P ( ) = R ( ) C ( ) = 30 (10, ) = , The graph of the profit function is: So the break even occurs when approimately 1,010 or 99,030 pairs of sandals are produced. Profit is maimized by selling 50,00 pairs of sandals.. We know that s() t = 4.9t + v0t+ s0 gives the position of a particle at a time t 0. Here v 0 = 5 and s 0 = 5, then; st () = 4.9t + 5t+ 5. The particle strikes the ground at the time t where st () = 4.9t + 5t+ 5= 0. This occurs when t 5.95, that is the particle stays in the air for approimately 5.95 seconds. 5 The particle is at the maimum height when t =.55. ( 4.9) And the maimum height that the particle reaches is : s (.55) = meters.