x² + 9x + 1
let's start by grouping the ones with the same variable
(x² + 9x) + 1
(x² + 9x + [?]²) + 1
so, we seem to be missing a value there, to get a perfect square trinomial, let's recall a perfect square trinomial has a middle term that is 2 * "other two values", namely
so, the middle term on this group will be 9x.
we know 2*x*[?] = 9x, so then
so that's our mystery felllow.
now, let's bear in mind that we'll be borrowing from our very good friend Mr Zero, 0, so if we add (9/2)², we also have to subtract (9/2)².
![\bf \left[x^2+9x+\left( \cfrac{9}{2} \right)^2 - \left( \cfrac{9}{2} \right)^2 \right]+1\implies \left[x^2+9x+\left( \cfrac{9}{2} \right)^2 \right]+1- \left( \cfrac{9}{2} \right)^2 \\\\\\ \left( x+\cfrac{9}{2} \right)^2+1-\cfrac{81}{4}\implies \left( x+\cfrac{9}{2} \right)^2-\cfrac{77}{4}](https://tex.z-dn.net/?f=%5Cbf%20%5Cleft%5Bx%5E2%2B9x%2B%5Cleft%28%20%5Ccfrac%7B9%7D%7B2%7D%20%5Cright%29%5E2%20-%20%5Cleft%28%20%5Ccfrac%7B9%7D%7B2%7D%20%5Cright%29%5E2%20%5Cright%5D%2B1%5Cimplies%20%5Cleft%5Bx%5E2%2B9x%2B%5Cleft%28%20%5Ccfrac%7B9%7D%7B2%7D%20%5Cright%29%5E2%20%5Cright%5D%2B1-%20%5Cleft%28%20%5Ccfrac%7B9%7D%7B2%7D%20%5Cright%29%5E2%20%5C%5C%5C%5C%5C%5C%20%5Cleft%28%20x%2B%5Ccfrac%7B9%7D%7B2%7D%20%5Cright%29%5E2%2B1-%5Ccfrac%7B81%7D%7B4%7D%5Cimplies%20%5Cleft%28%20x%2B%5Ccfrac%7B9%7D%7B2%7D%20%5Cright%29%5E2-%5Ccfrac%7B77%7D%7B4%7D)
<h2>
Answer:</h2><h2>40 minutes</h2>
Step-by-step explanation:
Answer:
The vertex is at (1, -108).
Step-by-step explanation:
We have the function:
And we want to find its vertex point.
Note that this is in factored form. Hence, our roots/zeros are <em>x</em> = 7 and <em>x</em> = -5.
Since a parabola is symmetric along its vertex, the <em>x-</em>coordinate of the vertex is halfway between the two zeros. Hence:
To find the <em>y-</em>coordinate, substitute this back into the function. Hence:
Therefore, our vertex is at (1, -108).
(x-3)(x-5)
i hope this helps :)
Answer:
-2(2x+3)
Step-by-step explanation:
