# chapter 4 section 2 graphing quadratic functions in vertex or intercept form

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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. 3. Calculate minimum or maximum value. - PowerPoint PPT PresentationTRANSCRIPT

Chapter 4 Section 2Graphing Quadratic Functions in Vertex or

Intercept Form

In this assignment, you will be able to...

1. Graph a quadratic function in the vertex orIntercept 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 noticethat 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 noticethat 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 noticethat 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)

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

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+3 Multiply 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 -0.001x=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)(130-

260) 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.)

You have finished 4-2