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.
Answer:
Step-by-step explanation:
The answer is 20.
Hope this helps!! May I have brainliest?
Answer: (3x + 11y)^2
Demonstration:
The polynomial is a perfect square trinomial, because:
1) √ [9x^2] = 3x
2) √121y^2] = 11y
3) 66xy = 2 *(3x)(11y)
Then it is factored as a square binomial, being the factored expression:
[ 3x + 11y]^2
Now you can verify working backwar, i.e expanding the parenthesis.
Remember that the expansion of a square binomial is:
- square of the first term => (3x)^2 = 9x^2
- double product of first term times second term =>2 (3x)(11y) = 66xy
- square of the second term => (11y)^2 = 121y^2
=> [3x + 11y]^2 = 9x^2 + 66xy + 121y^2, which is the original polynomial.
