Question

major math problem...help!!!
find the vertex, focus, and directrix of the paarabola and sketch it graph. (x+3/2)^2=4(y-2)
i have found the vertex and it is (-3/2,2)... i'm not sure how to find the others :(

Answer 1

rearrange the equation
(x+3/2)^2=4(y-2)
y=(x+3/2)^2/4+2

Answer 2

can u show step-by-step how u did that plz?

Answer 3

rearrange the equation
(x+3/2)^2=4(y-2)
divide 4 both sides
rearrange the equation
(x+3/2)^2/4=(y-2)
add 2 to both sides
y=(x+3/2)^2/4+2

Answer 4

then expand (x+3/2)^2,
\[y=\frac{(x+3/2)^2}{4}+2\]
\[y=\frac{x^2+3x+\frac{9}{4}}{4}+2\]
multiply 4
\[y=\frac{4x^2+12x+9}{16}+2\]
\[y=\frac{x^2}{4}+\frac{3 x}{4}+\frac{41}{16}\]

Answer 5

what is that above the 4 in the second equation???

Answer 6

9

Answer 7

o. :)

Answer 8

whats the next step?

Answer 9

the vertex i had was correct right?

Answer 10

yes

Answer 11

okay... then how could get the focus for this parabola?

Answer 12

use \[y=\frac{x^2}{4}+\frac{3 x}{4}+\frac{41}{16}\]
find the y-intercept

Answer 13

so i would set x to zero right?

Answer 14

yes

Answer 15

okay let me do it and u could tell me if im right

Answer 16

41/16?

Answer 17

yes

Answer 18

okay so the my focus is (-3/2, 41/16?

Answer 19

focus is
\[\left(-\frac{3}{2},3\right)\]

Answer 20

how is it 3???????

Answer 21

the complete the square form of
\[y=\frac{x^2}{4}+\frac{3 x}{4}+\frac{41}{16}\]

\[\frac{1}{4} \left(x+\frac{3}{2}\right)^2+2\]
p=1/4(1/4)=1

focus=( h, c+1)
focus=(-3/2 , 2+p)
focus=(-3/2 ,3)

Answer 22

?

Answer 23

?

Answer 24

lol no i need help w/ what u just said
or wrote

Answer 25

\[\frac{1}{4} \left(x+\frac{3}{2}\right)^2+2\]

\[a \left(x+b\right)^2+c\]
\[p=\frac{1}{4a}\]

\[focus=( b, c+p) \]

Answer 26

oka so that equation is used for what?

Answer 27

you mean p?

Answer 28

yeas

Answer 29

that's a formula you need to memorize

Answer 30

and what exactly does that formula solve for?

Answer 31

focus

Answer 32

ooooooooooooo really??? so that the formula to find the focus on all problem right?