Question

4.2 day 1
graph quadratic functions in vertex form ia
graph the function. label the vertex and the axis of symmetry. use your calculator to find 5 points on the parabola including the vertex. state the maximum or minimum value of the graph and compare the width of the graph to the width of $y = x^2$

1. $y = \frac{1}{2}(x - 3)^2 - 4$

• x y table (vertex indicated)
• 1. v : ( , )
• 2. aos: $x = $ ______
• 3. maximum or minimum?
• 4. value: _______________
• (graph grid)

1. $y = -(x - 1)^2 + 5$

• x y table (vertex indicated)
• 1. v : ( , )
• 2. aos: $x = $ ______
• 3. maximum or minimum?
• 4. value: _______________
• (graph grid)

Explanation:

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For Function 1: $y=\frac{1}{2}(x-3)^2 - 4$

Step1: Identify vertex form

Vertex form: $y=a(x-h)^2+k$, vertex $(h,k)$
Here $h=3$, $k=-4$, $a=\frac{1}{2}$

Step2: Find vertex (V)

Vertex is $(h,k)$
$V=(3, -4)$

Step3: Find axis of symmetry (AOS)

AOS is $x=h$
$x=3$

Step4: Determine max/min

$a=\frac{1}{2}>0$, so parabola opens up: Minimum

Step5: Find min value

Minimum value is $k=-4$

Step6: Calculate 5 points

• Vertex: $x=3$, $y=\frac{1}{2}(3-3)^2-4=-4$
• $x=1$: $y=\frac{1}{2}(1-3)^2-4=\frac{1}{2}(4)-4=2-4=-2$
• $x=2$: $y=\frac{1}{2}(2-3)^2-4=\frac{1}{2}(1)-4=-3.5$
• $x=4$: $y=\frac{1}{2}(4-3)^2-4=\frac{1}{2}(1)-4=-3.5$
• $x=5$: $y=\frac{1}{2}(5-3)^2-4=\frac{1}{2}(4)-4=-2$

Step7: Compare width to $y=x^2$

$|a|=\frac{1}{2}<1$, so parabola is wider than $y=x^2$

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For Function 2: $y=-(x-1)^2 + 5$

Step1: Identify vertex form

Vertex form: $y=a(x-h)^2+k$, vertex $(h,k)$
Here $h=1$, $k=5$, $a=-1$

Step2: Find vertex (V)

Vertex is $(h,k)$
$V=(1, 5)$

Step3: Find axis of symmetry (AOS)

AOS is $x=h$
$x=1$

Step4: Determine max/min

$a=-1<0$, so parabola opens down: Maximum

Step5: Find max value

Maximum value is $k=5$

Step6: Calculate 5 points

• Vertex: $x=1$, $y=-(1-1)^2+5=5$
• $x=-1$: $y=-(-1-1)^2+5=-(4)+5=1$
• $x=0$: $y=-(0-1)^2+5=-(1)+5=4$
• $x=2$: $y=-(2-1)^2+5=-(1)+5=4$
• $x=3$: $y=-(3-1)^2+5=-(4)+5=1$

Step7: Compare width to $y=x^2$

$|a|=1$, so parabola has same width as $y=x^2$

Answer:

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Function 1: $y=\frac{1}{2}(x-3)^2 - 4$

1. V: $(3, -4)$
2. AOS: $x=3$
3. Maximum or Minimum? Minimum
4. Value: $-4$
5. Table of 5 points:

X	Y
2	-3.5
3	-4
4	-3.5
5	-2

1. Width comparison: Wider than $y=x^2$

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Function 2: $y=-(x-1)^2 + 5$

1. V: $(1, 5)$
2. AOS: $x=1$
3. Maximum or Minimum? Maximum
4. Value: $5$
5. Table of 5 points:

X	Y
0	4
1	5
2	4
3	1

1. Width comparison: Same width as $y=x^2$

(Graphing note: Plot the table points, draw a smooth parabola through them, label the vertex and axis of symmetry on the grid for each function)