something noteworthy is that the independent and squared variable in this case will be the "x", namely the graph of that quadratic is a vertical parabola.
![\bf f(x) = (x+2)(x-4)\implies 0=(x+2)(x-4)\implies x = \begin{cases} -2\\ 4 \end{cases} \\\\\\ \boxed{-2}\rule[0.35em]{7em}{0.25pt}0\rule[0.35em]{3em}{0.25pt}\stackrel{\downarrow }{1}\rule[0.35em]{10em}{0.25pt}\boxed{4}](https://tex.z-dn.net/?f=%5Cbf%20f%28x%29%20%3D%20%28x%2B2%29%28x-4%29%5Cimplies%200%3D%28x%2B2%29%28x-4%29%5Cimplies%20x%20%3D%20%5Cbegin%7Bcases%7D%20-2%5C%5C%204%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20%5Cboxed%7B-2%7D%5Crule%5B0.35em%5D%7B7em%7D%7B0.25pt%7D0%5Crule%5B0.35em%5D%7B3em%7D%7B0.25pt%7D%5Cstackrel%7B%5Cdownarrow%20%7D%7B1%7D%5Crule%5B0.35em%5D%7B10em%7D%7B0.25pt%7D%5Cboxed%7B4%7D)
so the parabola has solutions at x = -2 and x = 4, and its vertex will be half-way between those two guys, namely at x = 1.
since this is a vertical parabola, its axis of symmetry, the line that splits its into twin sides, will be a vertical line, and it'll be the x-coordinate of the vertex, since the vertex hasa a coordinate of x = 1, then the axis of symmetry is the vertical line of x = 1.
I believe the answer is 16
The question is telling you that the length of the rectangle is 3 metres more than twice the width.
So let:
<em>w= width</em>
<em>L= length</em>
Because the length is 3 metres more than twice<em> </em>the width: <em>L= </em><em>2</em><em>w+</em><em>3</em>
They also tell you the perimeter is 48 metres.
<em>P= L+L+w+w</em>
So the equation of the perimeter is:
<em>48= (2w+3)+(2w+3)+2w +2w</em>
<em>48= 2(2w+3) + 4w</em>
To find w, expand and simplify.
<em>48= 4w+6+4w</em>
<em>48= 8w + 6</em>
<em>42= 8w</em>
<em>5.25=w</em>
Now that you know the width, plug in the value into the length equation:
<em>L= 2w+3</em>
<em>L=2(5.25)+3</em>
<em>L=10.50+3</em>
<em>L=13.5</em>
If I am wrong let me know! I hope this helps.