Question

1. $f(x) = -2x^2 - 4x + 3$

x	y
-1	5
0	3
1	-3
2	-13

1. v : ( , )
2. aos: $x = $ ______
3. maximum or minimum?
4. value: ______

1. $f(x) = \frac{1}{2}x^2 - 4x + 2$

x	y

1. v : ( , )
2. aos: $x = $ ______
3. maximum or minimum?
4. value: ______

Explanation:

First, we correct and solve problem 3, then solve problem 4:

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For problem 3: $f(x) = -2x^2 -4x +3$

Step1: Find vertex x-coordinate

The x-coordinate of the vertex of $ax^2+bx+c$ is $x=-\frac{b}{2a}$. Here $a=-2, b=-4$:
$x = -\frac{-4}{2(-2)} = \frac{4}{-4} = -1$

Step2: Find vertex y-coordinate

Substitute $x=-1$ into $f(x)$:
$f(-1) = -2(-1)^2 -4(-1)+3 = -2+4+3=5$

Step3: Identify AOS

Axis of Symmetry (AOS) is the vertex x-coordinate: $x=-1$

Step4: Max/Min check

Since $a=-2<0$, parabola opens downward, so it has a maximum.

Step5: Max value

The maximum value is the vertex y-coordinate: 5

Step6: Verify table values

• $x=-2$: $f(-2)=-2(4)-4(-2)+3=-8+8+3=3$
• $x=0$: $f(0)=0-0+3=3$
• $x=1$: $f(1)=-2-4+3=-3$
• $x=2$: $f(2)=-2(4)-4(2)+3=-8-8+3=-13$

Step1: Find vertex x-coordinate

For $ax^2+bx+c$, $x=-\frac{b}{2a}$. Here $a=\frac{1}{2}, b=-4$:
$x = -\frac{-4}{2(\frac{1}{2})} = \frac{4}{1} = 4$

Step2: Find vertex y-coordinate

Substitute $x=4$ into $f(x)$:
$f(4) = \frac{1}{2}(4)^2 -4(4)+2 = \frac{1}{2}(16)-16+2=8-16+2=-6$

Step3: Identify AOS

Axis of Symmetry is the vertex x-coordinate: $x=4$

Step4: Max/Min check

Since $a=\frac{1}{2}>0$, parabola opens upward, so it has a minimum.

Step5: Min value

The minimum value is the vertex y-coordinate: $-6$

Step6: Complete the table

Calculate $y$ for $x=2,3,4,5,6$:

• $x=2$: $f(2)=\frac{1}{2}(4)-4(2)+2=2-8+2=-4$
• $x=3$: $f(3)=\frac{1}{2}(9)-4(3)+2=4.5-12+2=-5.5$
• $x=4$: $f(4)=-6$ (vertex)
• $x=5$: $f(5)=\frac{1}{2}(25)-4(5)+2=12.5-20+2=-5.5$
• $x=6$: $f(6)=\frac{1}{2}(36)-4(6)+2=18-24+2=-4$

Answer:

(Corrected for Problem 3):

1. V: $(-1, 5)$
2. AOS: $x=-1$
3. Maximum
4. Value: $5$

Table values are correct as given.

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For problem 4: $f(x) = \frac{1}{2}x^2 -4x +2$