PRACTICE PROBLEMS AND WORKED OUT SOLUTIONS
ProblemStandard Form-
1.2x^2+4x+6 (Level 1-2) 2.2x^2-4x+2 (Level 1-2) 3.-3x^2+4x-4 (Level 3-4) 4. 2x^2+2x+4 (Level 3-4) 5.-3x^2+5x+7(Level 5-6) 6. 4x^2-6x-9 (Level 5-6) 7. Marcus is feeling lucky today. So he is going to jump off of a 9 ft roof into a pool of pickles. As a function of time, Marcus' height can be determined by the function h(t)= -16t^2+16t+17. t stands for the time in seconds. How long will it take Marcus to hit the pool of pickles? (Level 7-8) 8. John just built a new catapult and plans to try it out. He will be launched at 115 ft per second. Its height h after t seconds is given by the equation h(t)= -16t^2+115t. How long will it take for John to reach his maximum height after he is launched from the catapult? (Level 7-8) Vertex Form-
1. y= -1/2(x + 4)^2 +6, (Level 1-2) 2. y= (x + 3)^2 – 4, (Level 1-2) 3. y= -2(x- 2)^2 +3, (Level 3-4) 4. y= (x + 1)(x + 3) (Level 3-4) 5. y= 3(x -2)(x +6) (Level 3-4) 6. y = 2(x – 1.75)^2 + 3.25 (Level 5-6) 7. y = -.5(x – 3.5)^2 + 6 (Level 5-6) |
Solution worked outStandard Form-
1. a=2 b=4 c=6 x=-b/2a x=-(4)/2(2)= -4/4=-1 2(-1)^2-4+6=4 x=-1 y=4 This graph opens up because the A value is a positive number 2. a=2 b=-4 c=2 -b/2a -(-4)/2(2)= -8/2= -4 2(-4)^2-4(-4)+2= 32+16+2= 50 x=-4 y=50 This graph opens up because the A value is a positive number 3. a=-3 b=4 c=-4 x=-b/2a x=-(4)/2(-3)=-4/6= -2/3 -2(-2/3)^2+4(-2/3)-4= -28/3 x= -2/3 y= -28/3 This graph opens down because the A value is negative 4. a=2 b=2 c=4 x=-b/2a x=-(2)/2(2)=-2/4=-1/2 2(-1/2)^2+2(-1/2)+4= 1/4-2.5+4= 1 and 3/4 x=-1/2 y=1 and 3/4 This graph opens up because the A value is a positive number 5.a=-3 b=5 c=7 x=-b/2a x=-(5)/2(-3)= -5/-6= 5/6 -3(5/6)^2+5(5/6)+7= -1.33 x=5/6 y=1.33 This graph opens down because the A value is negative 6. a=4 b=-7 c=-8 x=-b/2a x=-(-7)/2(4)= -7/8 4(-7/8)^2-6(-7/8)-8= -27.25 x=-7/8 y=-27.25 This graph opens up because the A value is positive 7. Like any other standard form problems, we find our A, B, and C values for the first step. A=-16 B=16 C=9. then, we put it into this formula: x=-b/2a x=-(16)/2(-16)= -16/-32= 1/2 -16(1/2)^2+16(1/2)+17 -16(.25)+8+17= 21 The answer is 1/2 a second because it is a more logical answer. The graph opens down because the A value is negative 8. We et this problem up like we do any other standard form problems. Find the A, B, and C values. A=-16 B=115 C=0 Then we set up the formula: x=-b/2a and plug in our values x=-(115)/2(-16)= -115/-32= 3.6 -16(3.6)^2+115(3.6)= -92.36 It will take John 3.6 seconds to reach his maximum height. And also because -92.36 is not a logical answer The graph opens down because the A value is negative Vertex Form- 1. Step 1: Identify “a” and “h” a) -1/2, h) 4. Step 2: Plot the point (h,k) = (4, -1/2), then draw the axis of symmetry. Step 3: Evaluate the function for the x values. x= 0: y= -1/2(0 + 4)^2 + 6 = -2, x= 2: y= -1/2(2 + 4)^2 + 6 = -12. Plot the points (0,-2) and (2,-12) and their reflections in the axis of symmetry. Step 4 Draw the parabola. 2. Step 1: Identify “a” and “h” a) 1, h) -4. Step 2: Plot the point (h,k) = (-4, 1), then draw the axis of symmetry. Step 3: Evaluate the function for the x values. x= 0: y= (0 + 3)^2 -4 = 5, x= 2: y= (2 + 3)^2 – 4 =21. Plot the points (0,5) and (2,21) and their reflections in the axis of symmetry. Step 4 Draw the parabola. 3.Step 1: Identify “a” and “h” a) -2, h) 2. Step 2: Plot the point (h,k) = (2, -2), then draw the axis of symmetry. Step 3: Evaluate the function for the x values. x= 0: y= -2(0 + 2)^2 + 3 = -5, x= 2: y= -2(2 + 2)^2 +3 = -29. Plot the points (0, -5) and (2,-29) and their reflections in the axis of symmetry. Step 4 Draw the parabola. 4. Step 1: Identify the x-intercepts. Because p = -1 and q = -3, the x-intercepts occur at the points (-1, 0) and (-3, 0). Step 2: Find the coordinates of the vertex x= p + q/2= -1 -3/2= -2. y= 2(-2 + 1)(-2 + 3)= -2. So the vertex is (-2, -2). Step 3: Draw a parabola through the vertex and points where the x-intercepts occur. 5.Step 1: Identify the x-intercepts. Because p = 2 and q = -6, the x-intercepts occur at the points (2, 0) and (-6, 0). Step 2: Find the coordinates of the vertex x= p + q/2= 2 - 6/2= -2. y= 2(-2 + 1)(-2 + 3)= -2. So the vertex is (-2, -2). Step 3: Draw a parabola through the vertex and points where the x-intercepts occur. 6.Step 1: Identify “a” and “h” a) 2, h) -1.75. Step 2: Plot the point (h,k) = (-1.75, 2), then draw the axis of symmetry. Step 3: Evaluate the function for the x values. x= 0: y= 2(0 - 1.75)^2 + 3.25 = 9.375, x= 2: y= 2 (2 – 1.75)^2 + 3.25 = 4.375. Plot the points (0, 9.375) and (2, 4.375) and their reflections in the axis of symmetry. Step 4 Draw the parabola. 7. Step 1: Identify “a” and “h” a) .5, h) -3.5. Step 2: Plot the point (h,k) = (-3.5, .5), then draw the axis of symmetry. Step 3: Evaluate the function for the x values. x= 0: y= .5(0 -3.5)^2 + 6 = 12.125, x= 2: y= .5(2 – 3.5)^2 + 6 = 7.125. Plot the points (0, 12.125) and (2, 7.125) and their reflections in the axis of symmetry. Step 4 Draw the parabola. |