Question

Find the focus of the parabola y = ⅛(x - 3)2 + 1.

Select one:
a. (-3, -1)
b. (5, 1)
c. (-1, 3)
d. (3, 3)

Answer 1

(x-h)^2 = 4p(y-k), where p is the distance from the vertex to the focus

Answer 2

find the vertex, and find p, then you'll have the focus

Answer 3

SO is it C? @sourwing

Answer 4

what did you get for the vertex?

Answer 5

I just got a line not a parabola @jdoe0001

Answer 6

hmm that looks like a parabola -> \(\bf y=\cfrac{1}{8}(x-3)^2+1\)

Answer 7

can you tell the vertex from it?

Answer 8

ttp://www.mathwarehouse.com/geometry/parabola/images/standard-vertex-forms.png

Answer 9

shoot... missing letter =)
http://www.mathwarehouse.com/geometry/parabola/images/standard-vertex-forms.png

Answer 10

nevermind I got the parabola

Answer 11

3,1

Answer 12

is the vertex @jdoe0001

Answer 13

yeap
and "p" as @sourwing already pointed out, is the distance from the vertex to the focus

the parabola is "x" based, so is vertical

so we know the parabola has a POSITIVE leading term coefficient
that means the parabola is going UP

so from the vertex of (3,1) it goes up, meaning the "y" value moves

how far up? well, "p" distance

\(\bf (x-{\color{red}{ h}})^2=4{\color{green}{ p}}(y-{\color{blue}{ k}})\qquad vertex\quad ({\color{red}{ h}},{\color{blue}{ k}})
\\ \quad \\ \quad \\
y=\cfrac{1}{8}(x-3)^2+1\implies y-1=\cfrac{1}{8}(x-3)^2
\\ \quad \\
\implies 8(y-1)=(x-3)^2\)

can you tell "p" from there?

Answer 14

.5

Answer 15

well \(\bf (x-{\color{red}{ h}})^2=4{\color{green}{ p}}(y-{\color{blue}{ k}})\qquad vertex\quad ({\color{red}{ h}},{\color{blue}{ k}})
\\ \quad \\ \quad \\
\implies 8(y-1)=(x-3)^2\implies 4{\color{green}{ p}}=8\implies p=?\)

Answer 16

2 nevermind

Answer 17

yeap :), so the focus is at (3, 1 + 2 )

Answer 18

so 3,3?

Answer 19

yeap