Question

Help with graphing!!! (MEDAL)

Answer 1

@ybarrap

Answer 2

how do i figure out the equation?

Answer 3

y = (x-h)^2 + k

where (h,k) is the vertex

Answer 4

so for the first one, i would have (0,2) as the vertex right?

Answer 5

sorry (0,1)

Answer 6

The first one has a vertex of (0,8) ?

Answer 7

For #1 https://www.desmos.com/calculator/hymmkpclji

Answer 8

Using the information provided by @TheSmartOne :
1st Parabola - (h,k) = (0,9)
2nd Parabola - (h,k) = (0,1)
Equation of parabola
$$
y=a(x-h)^2 + k
$$
The constant \(a\) determines how wide and whether the parabola points up or down.

Answer 9

oh yes. how did u get that equation?

Answer 10

i need step by step help please

Answer 11

Here is a quick explanation
https://en.wikipedia.org/wiki/Parabola#Equations

Answer 12

You can also take the roots, or zeroes of the given graph and using the general form:
\[f(x)=(x-a_1)(x-a_2)...(x-a_n)\]
Where a1 to "an" are the roots of the function in question.

Answer 13

Where \(a=1/4p\) and p is the distance from the vertex to the focus.

But more importantly, the distance from the Focus F(p1,p2) to a point on the parabola, say, P(x,y) is the same as the distance from it to the directrix. "p" is also the distance from the vertex to the directrix:

$$
\sqrt{(x-h)^2+(y-k^2)}=\sqrt{(x+p)^2}
$$
If you solve this equation for y, you'll get
$$
y =\frac{(x-h)^2}{4p}+k\\
y=a(x-h)^2+k
$$
where \(a=1/4p\)