Lesson 4.1 Exercises, pages

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1 Lesson 4.1 Eercises, pages When approimating answers, round to the nearest tenth. A 4. Identify the y-intercept of the graph of each quadratic function. a) y = b) y = Use mental math. Substitute 0. y 1 y 3 c) y = d) y = 4 3 y 0 y 0 5. State whether the verte of the graph of each quadratic function is a maimum point or a minimum point. a) y = b) y = 5-3 Coefficient of is positive. So, parabola opens up. Verte is a minimum point. Coefficient of is negative. So, parabola opens down. Verte is a maimum point. 1

2 B 6. Identify whether each table of values represents a linear function, a quadratic function, or neither. Eplain how you know. a) b) y y The -coordinates decrease The -coordinates increase by 1 each time. by each time. First differences: First differences: ( 3) ( ) ( 7) ( 16) 11 The first differences are not The first differences decrease constant, and they do not increase by each time. So, the or decrease by the same number. function is quadratic. So the function is neither linear nor quadratic. 7. Use a table of values to graph each quadratic function below, for the values of indicated. Determine: i) the intercepts ii) the coordinates of the verte iii) the equation of the ais of symmetry iv) the domain of the function v) the range of the function a) y = y y y 1 3 i) -intercepts: 4, 8 from equation, y-intercept: 3 ii) verte: (6, 4) iii) ais of symmetry: 6 v) range: y 4, y ç 0 6 b) y = i) -intercepts: 3, 1 y-intercept: 6 ii) verte: (, ) iii) ais of symmetry: v) range: y», y ç y y 8 6 y DO NOT COPY. P

3 c) y = y i) -intercepts: 3, 5 from equation, y-intercept: 5 ii) verte: (4, 3) iii) ais of symmetry: 4 v) range: y 3, y ç y y a) Use a graphing calculator to graph each set of quadratic functions. i) y = + ii) y = - - y = y = + + y = y = b) How many -intercepts may a parabola have? y has -intercepts; y 1 has 1 -intercept; and y has no -intercepts. y has -intercepts; y 1 has 1 -intercept; and y has no -intercepts. So, a parabola may have 0, 1, or -intercepts. c) How many y-intercepts may a parabola have? Each of the quadratic functions in part a has 1 y-intercept. So, a parabola may have 1 y-intercept. 9. Use a graphing calculator to graph each quadratic function below. Determine: i) the intercepts ii) the coordinates of the verte iii) the equation of the ais of symmetry iv) the domain of the function v) the range of the function a) y = b) y = Once I had graphed a function, I used the CALC feature, where necessary, to determine the intercepts and the coordinates of the verte. When the intercepts were approimate, I wrote them to the nearest hundredth. i) -intercepts: none i) -intercepts: none y-intercept: 5 y-intercept: 15 ii) verte: (, 3) ii) verte: (4, 3) iii) ais of symmetry: iii) ais of symmetry: 4 v) range: y» 3, y ç v) range: y 3, y ç 3

4 c) y = d) y = i) -intercepts: about 0.66 i) -intercepts: about 1.17 and about.16 and about.33 y-intercept:.875 y-intercept: ii) verte: (0.75, ) ii) verte: (1.75, 1) iii) ais of symmetry: 0.75 iii) ais of symmetry: 1.75 v) range: y», y ç v) range: y 1, y ç 10. A stone is dropped from a bridge over the Peace River. The height of the stone, h metres, above the river, t seconds after it was dropped, is modelled by the equation h = 0-4.9t. a) Graph the quadratic function, then sketch it below. I input the equation y in my graphing calculator, then graphed the function. b) When did the stone hit the river? To the nearest tenth, the positive t-intercept is.0. So, the stone hit the river approimately s after it was dropped. c) What is the domain? What does it represent? The domain is the set of possible t-values; that is, all values between and including 0 and the positive t-intercept. To the nearest tenth of a second, the domain is: 0 t.0. The domain represents the time the stone was in the air. C 11. Consider the quadratic function y = a + c. What must be true about a and c in each case? a) The function has no -intercepts. When a is positive, the parabola opens up. So, for the function to have no -intercepts, its verte must be above the -ais; that is c>0. When a is negative, the parabola opens down. So, for the function to have no -intercepts, its verte must be below the -ais; that is c<0. 4 DO NOT COPY. P

5 b) The function has one -intercept. For the function to have one -intercept, the verte of the parabola must be on the -ais. So, c 0 c) The function has two -intercepts. When a is positive, the parabola opens up. So, for the function to have two -intercepts, its verte must lie below the -ais; that is c<0. When a is negative, the parabola opens down. So, for the function to have two -intercepts, its verte must lie above the -ais; that is c>0. 1. In Eample 3 on page 55, the parabola has one positive t-intercept which is approimately 1.3. The graph has also a negative t-intercept. a) Determine this intercept. Using the CALC feature on my graphing calculator, the negative t-intercept, to the nearest hundredth, is b) How could you interpret this intercept in terms of the situation modelled by the equation? The object is fired from a location just above the ground. The numerical value of the negative t-intercept could represent the additional time that it would take the object to reach its maimum height if it were fired from the ground. 5

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