Question

Find the focus of the parabola y = ½(x - 4)2 - 3

Select one:
a. (3.5, -3)
b. (4, -3.125)
c. (4, -2..875)
d. (4.5, -3)

Answer 1

@terenzreignz

Answer 2

x-coordinate is 4 -- obvious by inspection (axis of symmetry).

y-coordinate > -3 -- obvious by inspection (positive leading coefficient with vertex at (4,-3))

Answer 3

@terenzreignz

Answer 4

@TuringTest

Answer 5

@Luis_Rivera @zepdrix

Answer 6

How do you solve this @jdoe0001

Answer 7

\(\bf 4p(y-{\color{blue}{ k}})=(x-{\color{red}{ h}})^2\qquad
\begin{array}{llll}
vertex\ ({\color{red}{ h}}\quad ,\quad {\color{blue}{k}})\\
p=\textit{distance from the vertex to the focus}
\end{array}
\\-------------------------------
\\ \quad \\
y=\cfrac{1}{2}(x-4)^2-3\implies y+3=\cfrac{1}{2}(x-4)^2\implies (y+3)=\cfrac{(x-4)^2}{2}
\\ \quad \\
2(y+{\color{blue}{ 3}})=(x-{\color{red}{ 4}})^2\)

so.. what do you think "p" is?

Answer 8

What, thinking about the definition was too easy?

Answer 9

eheh

Answer 10

what is p?

Answer 11

well, what is it? notice the template and what you'd have \(\bf {\color{green}{ 2}}(y+{\color{blue}{ 3}})=(x-{\color{red}{ 4}})^2\\
{\color{green}{ 4p}}(y-{\color{blue}{ k}})=(x-{\color{red}{ h}})^2\)

what do you think would be "p"?

Answer 12

1

Answer 13

4p = 2

Answer 14

p=2

Answer 15

so find "p" and from the vertex, move over THAT MUCH to get the point of the focus

Answer 16

ok so 4,-1?

Answer 17

the vertex is 4,-3

Answer 18

hmmm what did you get for the vertex?

Answer 19

right so .... what's "p" ?

Answer 20

p is 2

Answer 21

4p = 2 so if you solve for "p" you get 2 ?

Answer 22

no nevermind .5

Answer 23

so the foucs is 3.5,-3

Answer 24

ok thanks a lot

Answer 25

yw

Answer 26

??

y = ½(x - 4)2 - 3

Vertex: (4,-3)
Axis of Symmetry: x = 4
x is quadratic and y is linear (Opens up or down)
1/2 > 0 (Opens up)
Focus: (4,-3+Something)
Directrix: y = -3-Something

How did the focus move from x = 4?