first off, let's notice the parabola is a vertical one, therefore the squared variable is the x, and the parabola is opening upwards, meaning the coefficient of x² is positive.
let's notice the vertex, or U-turn, is at (-2, 2)
![\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} \boxed{y=a(x- h)^2+ k}\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{-2}{ h},\stackrel{2}{ k}) \\\\\\ y=+1[x-(-2)]^2+2\implies y=(x+2)^2+2](https://tex.z-dn.net/?f=%20%5Cbf%20~~~~~~%5Ctextit%7Bparabola%20vertex%20form%7D%20%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%20%5Cboxed%7By%3Da%28x-%20h%29%5E2%2B%20k%7D%5C%5C%5C%5C%20x%3Da%28y-%20k%29%5E2%2B%20h%20%5Cend%7Barray%7D%20%5Cqquad%5Cqquad%20vertex~~%28%5Cstackrel%7B-2%7D%7B%20h%7D%2C%5Cstackrel%7B2%7D%7B%20k%7D%29%20%5C%5C%5C%5C%5C%5C%20y%3D%2B1%5Bx-%28-2%29%5D%5E2%2B2%5Cimplies%20y%3D%28x%2B2%29%5E2%2B2%20)
Answer:
-1
Step-by-step explanation:
See the attachment for the polynomial long division. The constant in the quotient is -1.
_____
Here, there is a remainder of -x. If there were no remainder the constant in the quotient is the ratio of the constant in the dividend to the constant in the divisor: -2/2 = -1.
That could be a first guess in a "guess and check" solution approach.
<em>Guess</em>: first term of binomial quotient is (2x^3)/x^2 = 2x; last term of binomial quotient is -2/2 = -1. So, the quotient is guessed to be (2x -1).
<em>Check</em>: (2x -1)(x^2 -x +2) = 2x^3 -3x^2 +5x -2
Subtracting this from the actual dividend gives a remainder of -x. This has a lower degree than the divisor, so no further adjustment of the quotient is required.
X=30.4-2y
(step by step) :
Answer:
Step-by-step explanation:
56y^2/9-30y+25
<u>Answer:</u>
x=3
<u>Step-by-step explanation:</u>
3(3 + x)=15+x
<u>distribute</u>
9 + 3x= 15+x
<u>subtract x from each side</u>
9+3x-x = 15
9+2x = 15
<u>subtract 9 from each side</u>
9+2x-9 = 15-9
2x = 6
<u>divide by 2</u>
2x/2 = 6/2
x=3