U4D3 Have out: Bellwork: total: Assignment, red pen, highlighter, textbook, GP notebook, calculator U4D3 Have out: Bellwork: Given f(x) = x2 + 10x – 24, answer the following: a) Algebraically solve for the x–intercepts. b) Determine the line of symmetry. c) Determine the vertex. d) Determine the y–intercept. e) Write the equation in vertex form. total:

Bellwork: total: Given f(x) = x2 + 10x – 24, answer the following: a) Algebraically solve for the x–intercepts. x2 + 10x – 24 = 0 +½ d) Determine the y–intercept. (x + 12)(x – 2) = 0 +½ +½ f(0) = (0)2 + 10(0) – 24 +1 x + 12 = 0 or x – 2 = 0 x = –12 x = 2 +½ f(0) = 0 + 0 – 24 (–12, 0) and (2, 0) +2 f(0) = –24 (0, –24) +1 b) Determine the line of symmetry. +1 = –5 e) Write the equation in vertex form. x = –5 +1 f(x) = (x + 5)2 – 49 +2 c) Determine the vertex. f(–5) = (–5)2 + 10(–5) – 24 +1 f(–5) = 25 – 50 – 24 total: f(–5) = –49 (–5, –49) +1

Equations of Parabolas Add to your notes: Equations of Parabolas Determine the equation of a parabola that passes through the given point and vertex. Steps: Example #1: point (0, –5), vertex (–2, 7) 1. Write the general vertex form equation of a parabola y = a(x – h)2 + k y = a(x + 2)2 + 7 2. Substitute the values for the vertex (h, k) and the other point (x, y) –5 = a(0 + 2)2 + 7 –5 = a(2)2 + 7 –5 = 4a + 7 3. Solve for the value of a. –7 –7 –12 = 4a 4. Write the new vertex form equation of a parabola 4 4 a = –3

Equations of Parabolas Add to your notes: Equations of Parabolas Determine the equation of a parabola that passes through the given point and vertex. Steps: Example #2: point (3, 4), vertex (1, –8) 1. Write the general vertex form equation of a parabola y = a(x – h)2 + k y = a(x – 1)2 – 8 2. Substitute the values for the vertex (h, k) and the other point (x, y) 4 = a(3 – 1)2 – 8 4 = a(2)2 – 8 4 = 4a – 8 3. Solve for the value of a. + 8 + 8 12 = 4a 4. Write the new vertex form equation of a parabola 4 4 a = 3

Finish the worksheet later. Determine the equation of a parabola that passes through the given point and vertex. Take out the worksheet: a) point (6, –2), vertex (1, 9) b) point (2, 0), vertex (–4, –7) y = a(x – h)2 + k y = a(x – h)2 + k y = a(x – 1)2 + 9 y = a(x + 4)2 – 7 –2 = a(6 – 1)2 + 9 0 = a(2 + 4)2 – 7 Finish the worksheet later. –2 = a(5)2 + 9 0 = a(6)2 – 7 –2 = 25a + 9 0 = 36a – 7 –9 –9 +7 +7 –11 = 25a 7 = 36a 25 25 36 36

a) Where should we put the x– and y–axes on the diagram? PG – 32 Parabolas are good models for all sorts of things in the world. Indeed, many animals and insects jump in parabolic paths. This diagram shows a jackrabbit jumping over a three foot high fence. In order to just clear the fence, the rabbit must start its jump at a point four feet from the fence. y (4, 3) vertex a) Where should we put the x– and y–axes on the diagram? The problem will be “nicer” to interpret if we center the axes at the left of the diagram. x (0, 0) (8, 0) Now that you have placed the axes, label all the important points that you notice.

Substitute the vertex (4, 3). y (4, 3) vertex y = a(x – 4)2 + 3 PG – 32 b) Find the equation of the parabola that models the jackrabbit’s jump. y = a(x – h)2 + k Substitute the vertex (4, 3). y (4, 3) vertex y = a(x – 4)2 + 3 Substitute one of the other points on the parabola. (0, 0) 0 = a(0 – 4)2 + 3 x Solve for a. (0, 0) (8, 0) 0 = a(– 4)2 + 3 0 = 16a + 3 –3 –3 –3 = 16a 16 16

The dependent variable represents the height. x PG – 32 c) SKIP! d) What do the dependent and independent variables represent in this situation? y The independent variable represents the position, or horizontal distance. (4, 3) vertex The dependent variable represents the height. x (0, 0) (8, 0) e) What are the domain and range of your equation? What parts of the domain and range are appropriate for the actual situation? For the equation: “Real life”: Domain: (–, ) or {x| x  R} Domain: [0, 8] or {x| 0  x  8} Range: (–, 3] or {y| – < y  3} Range: [0, 3] or {y| 0  y  3}

Substitute the vertex (60, 50). PG – 33 A fireboat in the harbor is assisting in putting out a fire in a warehouse along the pier. Use the same process as in the previous problem to find the equation of the parabola that models the path of the water from the fireboat to the fire. y x y = a(x – h)2 + k (60, 50) vertex Substitute the vertex (60, 50). y = a(x – 60)2 + 50 (120, 0) Substitute one of the other points on the parabola. (0, 0) (0, 0) 0 = a(0 – 60)2 + 50 Solve for a. 0 = a(– 60)2 + 50 0 = 3600a + 50 –50 –50 –50 = 3600a 3600 3600

g) Check your predictions with a graphing calculator. PG – 34 Make predictions about how many places the graph of each equation below will touch the x–axis. You may want to first rewrite some of the equations in a more useful form. (HINT: factor each equation and use the Zero Product Property!) a) y = (x – 2)(x – 3) b) y = (x + 1)2 c) y = x2 + 6x + 9 d) y = x2 + 7x + 10 e) y = x2 + 6x + 8 f) y = –x2 – 4x – 4 g) Check your predictions with a graphing calculator. h) Write a clear explanation describing how you can tell whether the equation of a parabola will touch the x–axis at only one point.

The parabola crosses 2 TIMES since there are 2 x–intercepts. PG – 34 a) y = (x – 2)(x – 3) b) y = (x + 1)2 c) y = x2 + 6x + 9 (x – 2)(x – 3) = 0 (x + 1)2 = 0 (x + 3)(x + 3) = 0 x – 2 = 0 or x – 3 = 0 x + 1 = 0 x + 3 = 0 x = 2 or x = 3 x = –1 x = –3 (2, 0) and (3, 0) (–1, 0) (–3, 0) The parabola crosses 2 TIMES since there are 2 x–intercepts. The parabola touches only ONCE since there is 1 x–intercept. The parabola touches only ONCE since there is 1 x–intercept.

The parabola crosses 2 TIMES since there are 2 x–intercepts. PG – 34 d) y = x2 + 7x + 10 e) y = x2 + 6x + 8 f) y = –x2 – 4x – 4 x2 + 7x + 10 = 0 x2 + 6x + 8 = 0 –x2 – 4x – 4 = 0 (x + 2)(x + 5) = 0 (x + 2)(x + 4) = 0 –(x2 + 4x + 4) = 0 x + 2 = 0 or x + 5 = 0 x + 2 = 0 or x + 4 = 0 –(x + 2)(x + 2) = 0 x = –2 or x = –5 x = –2 or x = –4 x + 2 = 0 (–2, 0) and (–5 , 0) (–2, 0) and (–4 , 0) x = –2 (–2, 0) The parabola crosses 2 TIMES since there are 2 x–intercepts. The parabola crosses 2 TIMES since there is 2 x–intercepts. The parabola touches only ONCE since there is 1 x–intercept. g) Check your predictions with a graphing calculator. I am right!  h) Write a clear explanation describing how you can tell whether the equation of a parabola will touch the x–axis at only one point. If the factored version includes a perfect square binomial factor, then the parabola will touch at one point only.

a) In the equation y = 2x + 5, what does the 2 tell you about graph? PG – 35 Draw accurate graphs of y = 2x + 5, y = 2x2 + 5, and y = ½x2 + 5 on the same set of axes. Label the intercepts. 5 –5 10 15 x y a) In the equation y = 2x + 5, what does the 2 tell you about graph? The 2 is the slope of the line. b) Is the 2 in y = 2x2 + 5 also the slope? Explain. The 2 is not the slope because only lines have CONSTANT slopes. The two in this case is a called a “STRETCH FACTOR.”

PG – 36 Do the sides of a parabola ever curve back in like the figure below? Give a reason for your answer. No. Parabolas do not curve back because as the x–values become larger (or smaller), the y–values become larger as well. Besides…recall the vertical line test. Parabolas are functions, and this means that they pass the vertical line test. (There is only one output for each input.)

No. Parabolas do not have any asymptotes. PG – 37 Do the sides of the parabola approach straight vertical lines as show below? (In other words, do parabolas have asymptotes?) Give a reason for your answer. No. Parabolas do not have any asymptotes. Vertical asymptotes suggest that there are x–values that cannot work in the equation of a parabola. However, every parabola has a domain of ALL REAL NUMBERS. Therefore, parabolas continue to widen forever to the left and right.

Equations of Parabolas WS Finish today's assignment: FX 32 - 45 Equations of Parabolas WS Be sure to cut out the tiles.