Question

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

x	y
-2
0
2
4

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

(grid for graphing)

1. \\( f(x) = 4x^2 + 1 \\)

x	y
-1
0
1
2

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

(grid for graphing)

Explanation:

Step1: Calculate y for $y=-\frac{1}{4}x^2$

For $x=-4$: $y=-\frac{1}{4}(-4)^2=-\frac{1}{4}(16)=-4$
For $x=-2$: $y=-\frac{1}{4}(-2)^2=-\frac{1}{4}(4)=-1$
For $x=0$: $y=-\frac{1}{4}(0)^2=0$
For $x=2$: $y=-\frac{1}{4}(2)^2=-\frac{1}{4}(4)=-1$
For $x=4$: $y=-\frac{1}{4}(4)^2=-\frac{1}{4}(16)=-4$

Step2: Find axis of symmetry for $y=-\frac{1}{4}x^2$

For $y=ax^2+bx+c$, axis is $x=-\frac{b}{2a}$. Here $b=0$, so $x=0$.

Step3: Calculate y for $f(x)=4x^2+1$

For $x=-2$: $f(-2)=4(-2)^2+1=4(4)+1=17$
For $x=-1$: $f(-1)=4(-1)^2+1=4(1)+1=5$
For $x=0$: $f(0)=4(0)^2+1=0+1=1$
For $x=1$: $f(1)=4(1)^2+1=4(1)+1=5$
For $x=2$: $f(2)=4(2)^2+1=4(4)+1=17$

Step4: Find axis of symmetry for $f(x)=4x^2+1$

For $f(x)=ax^2+bx+c$, $b=0$, so axis is $x=0$.

Answer:

For $y=-\frac{1}{4}x^2$:

X	Y
-2	-1
0	0
2	-1
4	-4

Axis of symmetry: $x=0$

For $f(x)=4x^2+1$:

X	Y
-1	5
0	1
1	5
2	17

Axis of symmetry: $x=0$

(Graphing note: For $y=-\frac{1}{4}x^2$, plot the points and draw a downward-opening parabola with vertex at (0,0). For $f(x)=4x^2+1$, plot the points and draw an upward-opening parabola with vertex at (0,1).)