To start, we're given the range that x lies in: from -1 to 4. We know from the fact that
![-1 \leq x](https://tex.z-dn.net/?f=-1%20%5Cleq%20x)
that -1 will be <em /><em>included</em> in that range, so we mark -1 on the number line with a solid circle. We also know from
![x\ \textless \ 4](https://tex.z-dn.net/?f=x%5C%20%5Ctextless%20%5C%204)
that, while x can be any value <em>up to</em> 4, it does not <em>include </em>4. We indicate this by drawing a hollow circle around 4 on the number line. Since x can be <em>any value within this range</em>, we make that fact clear by drawing a bold line between the points -1 and 4 on the number line. I've attached an image of what our final graph would look like.
Hello from MrBillDoesMath!
Answer:
4( x + 1.5)^2 + 0
Discussion:
4x^2 + 12x + 9 = => factor "4" from first 2 terms
4 (x^2 + 3x) + 9 = => complete the square, add\subtract (1.5)^2
4(x^2 + 3x + (1.5)^2) - 4 (1.5)^2 + 9 =
4 ( x + 1.5)^2 + ( 9 - 4(1.5)^2) = => as (1.5)^2 = 2.25
4 ( x + 1.5)^2 + ( 9 - 4(2.25)) = => as 4 ( 2.25) = 9
4 ( x+ 1.5)^2 + 0
Thank you,
MrB
The best and most correct answer among the choices provided by the question is the first choice. The first step to do during the substitution method is to s<span>olve the top equation for x. </span><span>I hope my answer has come to your help. God bless and have a nice day ahead!</span>
Answer:
Here's wh as t I found
(X-1)²+(y+4)²=4
Step-by-step explanation:
(x-h)²+(y-k)²=r² with the center being at the point (h,k)
and radius being r.
(x-1)²+(y-(-4))²=2²
(x-1)²+(y+4)²=4