Chapter 4 Section 2 Graphing Quadratic Functions in Vertex or Intercept Form In this assignment, you will be able to Graph a quadratic function in the vertex or Intercept Form. 2. Change Vertex or Intercept Form to Standard Form. 4. Calculate the height and distance of a jump. 3. Calculate minimum or maximum value.

1.) Graph the function. Label the vertex and axis of symmetry. y=(x-3)^2

Answer:y=(x-3)^2 If you look at Vertex Form y=a(x-h)+k, you will notice that h=3 and k=0. So your vertex is (3,0). Now make a T-Chart and pick 2 points above and below the vertex and solve for y..

Graph the function. Label the vertex and axis of symmetry. 2.)y=-(x+4)^2

Answer:y=-(x+4)^2 If you look at Vertex Form y=a(x-h)+k, you will notice that h=-4 and k=0. So your vertex is (-4,0). Now make a T-Chart and pick 2 points above and below the vertex and solve for y.

Graph the function. Label the vertex and axis of symmetry. 3.)y=2(x+1)^2-3

Answer: If you look at Vertex Form y=a(x-h)+k, you will notice that h=-1 and k=-3. So your vertex is (-1,-3). Now make a T-Chart and pick 2 points above and below the vertex and solve for y. y=2(x+1)^2-3

Graph the function. Label the vertex and axis of symmetry. 4.)y=-2(x-1)^2+1

Answer: If you look at Vertex Form y=a(x-h)+k, you will notice that h=1 and k=1. So your vertex is (1,1). Now make a T-Chart and pick 2 points above and below the vertex and solve for y. y=-2(x-1)^2+1

Graph the function. Label the vertex, axis of symmetry and x-intercepts. 5.)y=(x+2)(x+4)

If you take Intercept Form y=a(x-p)(x-q), you need to set each of the factors equal to zero. So x+2=0 and x+4=0. Then your x-intercepts are x=-2 and x=-4. Now find the point in the middle of the intercepts, x=-3. Answer:y=(x+2)(x+4) That is your axis of symmetry, x=-3. Now plug it in, y=(-3+2)(-3+4) or y=- 1. Vertex (-3,-1).

Graph the function. Label the vertex, axis of symmetry and x-intercepts. 6.)y=2(x-1)(x-5)

If you take Intercept Form y=a(x-p)(x-q), you need to set each of the factors equal to zero. So x-1=0 and x-5=0. Then your x-intercepts are x=1 and x=5. Now find the point in the middle of the intercepts, x=3. Answer: That is your axis of symmetry, x=3. Now plug it in, y=2(3-1)(3-5) or y=-8. Vertex (3,-8). y=2(x-1)(x-5)

Graph the function. Label the vertex, axis of symmetry and x-intercepts. 7.)y=-3x(x+8)

If you take Intercept Form y=a(x-p)(x-q), you need to set each of the factors equal to zero. So -3x=0 and x+8=0. Then your x-intercepts are x=0 and x=-8. Now find the point in the middle of the intercepts, x=-4. Answer: That is your axis of symmetry, x=-4. Now plug it in, y=-3*(-4)(-4+8) or y=48. Vertex (-4,48). y=-3x(x+8)

8.) First, identify the intercepts. Second, find the vertex. Calculate the minimum or maximum value. Then write the equation in Standard Form y=(x-4)(x-2)

Answer: x-intercepts (4,0) and(2,0) Vertex (3,-1) Minimum Value y=-1 Original Equation FOIL-Multiply Combine like x-terms y=x^2-4x-2x+8 y=x^2-6x+8

9.)y=-3(x-3)(x+2) First, identify the intercepts. Second, find the vertex. Calculate the minimum or maximum value. Then write the equation in Standard Form

y=-3(x-3)(x+2) Answer: Original Equation y=-3(x^2+2x-3x-6)FOIL-Multiply y=-3(x^2-x-6)Combine like x-terms Multiply parenthesis by -2y=-3x^2+3x+18 x-intercepts (3,0) and(-2,0) Vertex (1/2,18 3/4) Minimum Value y=18 3/4

10.) First, identify the vertex. Second, calculate the minimum or maximum value. Then write the equation in Standard Form. y=(x-2)^2+6

Answer: y=(x-2)(x-2)+6 y=(x^2-2x-2x+4)+6 y=(x^2-4x+4)+6 y=x^2-4x+10 Write out the squares FOIL-Multiply Combine like x-terms Combine 4+6 Original Equation Vertex (2,6) Minimum value of y=6

11.)y=-2(x+1)^2+3 First, identify the vertex. Second, calculate the minimum or maximum value. Then write the equation in Standard Form.

Answer: y=-2(x+1)(x+1)+3 y=-2(x^2+1x+1x+1)+3 y=-2(x^2+2x+1)+3 y=-2x^2-4x+1 Write out the squares FOIL-Multiply Combine like x-terms Combine -2+3 y=-2(x+1)^2+3 y=-2x^2-4x-2+3Multiply parenthesis by -2 Original Equation Vertex (-1,3) Maximum value of y=3

12.) Biology. The function y=-0.03(x-14)^2+6 models the jump of a red kangaroo where x is the horizontal distance (in feet) and y is the corresponding height (in feet). What is the kangaroo's maximum height? How long is the kangaroo's height?

Answer: In the function y=-0.03(x-14)^2+6, the vertex is (14,6). Therefor the kangaroo jumped to the height of the y coordinate or 6 feet high. Since the x-coordinate is at the half way point of the graph, the distance the kangaroo jumped is 2*14, or 28 feet.

13.) Golf. The flight of a particular golf shot can be modeled by the function y=-0.001x(x-260) where x is the horizontal distance (in yards) from the impact point and y is the height(in yards). The graph is below. a.)How many yards away from the impact point does the golf ball land? b.) What is the maximum height in yards of the golf shot?

Answer: First find the intercepts in the equation y=-0.001x(x-260) by setting x=0 and x-260=0. So x=0 and x=260. Therefore the ball travelled from o to 260 or 260 yards. At the half way point, 130 yards, the ball is at it's maximum height. So plug 130 in y=-0.001x(x-260) and you get y=-0.001(130)( ) or y=16.9 yards.

On a separate piece of paper, graph the equation y=-3(x-5)^2-4. Label the vertex and axis of symmetry. Describe whether the graph has a minimum or maximum value and calculate that value. Show all work and explanations. 14.)

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