Answer:
okay so I can't see the last option on ur sheet but when u graph this equation the point that lies on the curve is (2,-3) so that is the vertex but if I move up this curved line it passes through the Y axis at (0,-1), the Last answer that I can't see it wouldn't happen to be (-1,6) would it
Answer:
477.09 mi/da
Step-by-step explanation:
If student must use conversion given, 3.28ft = 1m, then the following works:
32km/hr = ______mi/da 1 day = 24 hrs, 1m=3.28ft, 1km=1000m,1mi=5280ft
32km/hr x 24hr/day x 1000m/1km x 3.28ft/1m x 1mi/5280ft = 477.09mi/da
If student can use 1km = 0.62mi, some steps can be skipped.
Tables are easy to make when u have the equation. Just simply sub any number in for x and solve for y
y = 3x + 2...lets let x = 0
y = 3(0) + 2
y = 2.....so when x = 0, y = 2....(0,2)
y = 3x + 2....lets let x = 1
y = 3(1) + 2
y = 3 + 2
y = 5....so when x = 1, y = 5....(1,5)
y = 3x + 2.....let x = 2
y = 3(2) + 2
y = 6 + 2
y = 8....so when x = 2, y = 8...(2,8)
u can find more point if u need to...
now make a table..
x y
0 2
1 5
2 8
and like I said, if u need more numbers for ur table, just do like I did above :)
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)
