a vertical axis, I assume it means a vertical axis of symmetry, thus it'd be a vertical parabola, like the one in the picture below.
![\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} y=a(x- h)^2+ k\qquad \qquad \leftarrow vertical\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{}{ h},\stackrel{}{ k}) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=0\\ k=0 \end{cases}\implies y=a(x-0)^2+0 \\\\\\ \textit{we also know that } \begin{cases} x=-2\\ y=3 \end{cases}\implies 3=a(-2-0)^2+0\implies 3=4a \\\\\\ \cfrac{3}{4}=a~\hspace{10em}y=\cfrac{3}{4}(x-0)^2+0\implies \boxed{y=\cfrac{3}{4}x^2}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%5Ctextit%7Bparabola%20vertex%20form%7D%20%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%20y%3Da%28x-%20h%29%5E2%2B%20k%5Cqquad%20%5Cqquad%20%5Cleftarrow%20vertical%5C%5C%5C%5C%20x%3Da%28y-%20k%29%5E2%2B%20h%20%5Cend%7Barray%7D%20%5Cqquad%5Cqquad%20vertex~~%28%5Cstackrel%7B%7D%7B%20h%7D%2C%5Cstackrel%7B%7D%7B%20k%7D%29%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cbegin%7Bcases%7D%20h%3D0%5C%5C%20k%3D0%20%5Cend%7Bcases%7D%5Cimplies%20y%3Da%28x-0%29%5E2%2B0%20%5C%5C%5C%5C%5C%5C%20%5Ctextit%7Bwe%20also%20know%20that%20%7D%20%5Cbegin%7Bcases%7D%20x%3D-2%5C%5C%20y%3D3%20%5Cend%7Bcases%7D%5Cimplies%203%3Da%28-2-0%29%5E2%2B0%5Cimplies%203%3D4a%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B3%7D%7B4%7D%3Da~%5Chspace%7B10em%7Dy%3D%5Ccfrac%7B3%7D%7B4%7D%28x-0%29%5E2%2B0%5Cimplies%20%5Cboxed%7By%3D%5Ccfrac%7B3%7D%7B4%7Dx%5E2%7D)
The system should look like this:
eh + b = 243
eh - b = 109
The answer will be 7 bean bags
Answer:
hey, there!
Step-by-step explanation:
The given is a point (-6,8) through which a line passes. And is perpendicular to the line y = 2x-4
The equation for point (-6,8) is,
(y-8)= m1(x+6)...........(i)
and given equation is y = 2x-4............(ii)
Now, from equation (ii).
slope (m2)= 2 { as equation (ii) is in the form of y= mx+c where m is a slope}.
Now, For perpendicular,
m1×m2= -1
m1×2= -1
Therefore, m1 = -1/2.
Putting, the value of m1 in equation (i).
(y-8) = -1/2×(x+6)
2(y-8)= -1(x+6)
2y - 16 = -x -6
x+2y-10 = 0......... is the required equation.
Hope it helps...
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