Question

given the table below write the equation of the quadratic in vertex form then convert it to standard form.

x	-4	-2	0	2	4
f(x)	2	0	2	8	18

vertex form:
*write the equation in the form y=a(x-h)^2+k without spaces.

standard form:
*write the equation in the form y=ax^2+bx+c without spaces.

Explanation:

Step1: Find the vertex

The vertex of a parabola is at the axis of symmetry. Looking at the table, when \( x = -2 \) and \( x = 0 \), the function values are symmetric around \( x=-1 \)? Wait, no, let's check the symmetry. The values of \( f(x) \) at \( x = -4 \) and \( x = 0 \) are both 2, at \( x = -2 \) is 0, \( x = 2 \) is 8, \( x = 4 \) is 18. Wait, actually, the vertex is at \( (-1, k) \)? Wait, no, let's see the symmetry. The points \( ( - 4,2) \) and \( (0,2) \) are symmetric about \( x=\frac{-4 + 0}{2}=-2 \)? Wait, no, \( f(-4)=2 \), \( f(0)=2 \), so the axis of symmetry is \( x=\frac{-4 + 0}{2}=-2 \)? Wait, no, \( x=-4 \) and \( x = 0 \): midpoint is \( x=\frac{-4+0}{2}=-2 \). And \( f(-2)=0 \), so the vertex is at \( (-2,0) \). So \( h=-2 \), \( k = 0 \). So vertex form is \( y=a(x + 2)^2+0=a(x + 2)^2 \).

Step2: Find the value of \( a \)

We can use another point, say \( x = 0 \), \( f(0)=2 \). Substitute into \( y=a(x + 2)^2 \): \( 2=a(0 + 2)^2=4a \), so \( a=\frac{2}{4}=\frac{1}{2} \). Wait, but let's check \( x = 2 \): \( y=\frac{1}{2}(2 + 2)^2=\frac{1}{2}(16)=8 \), which matches \( f(2)=8 \). \( x = 4 \): \( y=\frac{1}{2}(4 + 2)^2=\frac{1}{2}(36)=18 \), which matches \( f(4)=18 \). \( x=-4 \): \( y=\frac{1}{2}(-4 + 2)^2=\frac{1}{2}(4)=2 \), which matches. So \( a=\frac{1}{2} \). So vertex form is \( y=\frac{1}{2}(x + 2)^2 \).

Step3: Convert to standard form

Expand \( y=\frac{1}{2}(x + 2)^2 \). \( (x + 2)^2=x^2+4x + 4 \), so \( y=\frac{1}{2}(x^2+4x + 4)=\frac{1}{2}x^2+2x + 2 \).

Vertex Form:

\( y=\frac{1}{2}(x + 2)^2 \)

Standard Form:

\( y=\frac{1}{2}x^2+2x + 2 \)

Answer:

(Vertex Form):
\( y=\frac{1}{2}(x+2)^2 \)