Question

1. \\( y = -4x^2 + 8x + 2 \\)

x	y
-1	?
0	?
1	?
2	?
3	?

vertex →
axis of symmetry: \\( x = \underline{\quad} \\)

1. \\( f(x) = \frac{1}{2}x^2 + x - 3 \\)

x	y
-4
-3
-1
0
2

vertex →
axis of symmetry: \\( x = \underline{\quad} \\)

Explanation:

For Problem 3: $y=-4x^2+8x+2$

Step1: Verify given x-values

First, check the provided y-values by substituting each $x$ into the equation:

• For $x=-1$: $y=-4(-1)^2+8(-1)+2=-4-8+2=-10$ (the given $y=2$ is incorrect)
• For $x=0$: $y=-4(0)^2+8(0)+2=2$ (given $y=2$ is correct)
• For $x=1$ (vertex): $y=-4(1)^2+8(1)+2=-4+8+2=6$ (given $y=2$ is incorrect)
• For $x=2$: $y=-4(2)^2+8(2)+2=-16+16+2=2$ (given $y=0$ is incorrect)
• For $x=3$: $y=-4(3)^2+8(3)+2=-36+24+2=-10$ (given $y=0$ is incorrect)

Step2: Confirm axis of symmetry

For quadratic $ax^2+bx+c$, axis is $x=-\frac{b}{2a}$. Here $a=-4, b=8$:
$x=-\frac{8}{2(-4)} = 1$ (given axis is correct)

Step3: Correct coordinate table

$x$	$y=-4x^2+8x+2$
$0$	$2$
$1$	$6$
$2$	$2$
$3$	$-10$

Step1: Find vertex y-value

Vertex at $x=-1$ (given axis $x=-1$). Substitute $x=-1$:
$f(-1)=\frac{1}{2}(-1)^2+(-1)-3=\frac{1}{2}-1-3=-\frac{7}{2}=-3.5$

Step2: Calculate other y-values

• For $x=-4$: $f(-4)=\frac{1}{2}(-4)^2+(-4)-3=8-4-3=1$
• For $x=-2$: $f(-2)=\frac{1}{2}(-2)^2+(-2)-3=2-2-3=-3$
• For $x=0$: $f(0)=\frac{1}{2}(0)^2+0-3=-3$
• For $x=2$: $f(2)=\frac{1}{2}(2)^2+2-3=2+2-3=1$

Step3: Compile coordinate table

$x$	$f(x)$
$-2$	$-3$
$-1$	$-3.5$
$0$	$-3$
$2$	$1$

Answer:

• Corrected coordinate table:

$x$	$y$
$0$	$2$
$1$	$6$
$2$	$2$
$3$	$-10$

• Axis of symmetry: $x=1$
• The graph is a downward-opening parabola with vertex $(1,6)$, passing through the corrected points.

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For Problem 4: $f(x)=\frac{1}{2}x^2+x-3$