A parabola with vertex $(h,k)$ can be written in vertex form as

Since the vertex is $(3,-4)$, the equation must have the form

Use the point $(1,8)$ to find $a$:

So the equation is

The table shows that $f(-1)=8$ and $f(3)=8$, so the parabola has equal outputs for inputs that are 2 units from $x=1$. That means the axis of symmetry is $x=1$. Also, the table shows $f(1)=0$, so the vertex is $(1,0)$. A quadratic with vertex $(1,0)$ has the form $f(x)=a(x-1{)}^2$. Using the point $(-1,8)$:

so $a=2$. Therefore,

A quadratic has exactly one real solution when its discriminant is $0$. For $f(x)=x^2-6x+k$, the equation $x^2-6x+k=0$ has discriminant

Setting this equal to $0$ gives

so

Therefore, $k=9$ must be true.

Because the walkway is built inside all four sides, the dimensions of the planting region are reduced by $2x$ in each direction. Its dimensions are therefore $(10-2x)$ and $(4.8-2x)$. Set the area equal to $21.12$:

Expanding gives

Dividing by $0.8$ gives

Factoring gives

So $x=0.6$ or $x=11.2$. Since the width must fit inside the garden, $x=11.2$ is impossible. Therefore, the value of $x$ is $0.6$.

First use the area:

Applying the difference of squares gives

so

and, since $x>2$,

The dimensions are then $12$ feet by $8$ feet. The perimeter is

feet. At \$5perfoot,thetotalcostis$