Question

Help with finding the vertex of parabolas!

Answer 1

\[\large \frac{1}{2}(y + 4) = (x - 7)^2 \]

Answer 2

@Nnesha

Answer 3

Expand the equation first and solve for y to get the equation in standard form.

Answer 4

\[\frac{1}{2}y + 2 = x^2 - 14x + 49\]

Answer 5

actually, you don't need to rewrite the equation at all

Answer 6

the standard form is (x-h)^2=4p(y-k) where (h,k) is the vertex

Answer 7

well that's the vertex form you will see that n conics unit
but in algebra we use y=a(x-h)^2+k as vertex form

Answer 8

i already have (x - h)^2 + with (x-7)^2. how would i find out the value of k?

Answer 9

what number is replaced by h and k ??

Answer 10

h = 7?

Answer 11

k = 2 ?

Answer 12

\[\rm \large \frac{1}{2}(y + 4) = (x - 7)^2 \]
compare this with the equation ive posted above

Answer 13

k=2 ???

Answer 14

i don't understand

Answer 15

here is an example \[\large\rm (x-h)^2=4p(y-k)\] where(h,k) is the vertex
remember both h and k are positive
here is an example \[\rm (x-\color{ReD}{5})^2=4(y+\color{blue }{3})\] where (5 ,-3) is the vertex

Answer 16

do you know how to find h and k value from this equation \[\rm y=3(x-4)^2+5\] ?

Answer 17

would the h = 4 and k = 5?

Answer 18

yes right
that's how you have to find h and k value from this equation \[\rm (x-h)^2 =4p (y-k)\]

Answer 19

what does the p do in 4p?

Answer 20

if you have the equation of the parabola you just need to somplete the square

Answer 21

if you know the completing the square method, then you are ready tackle these.

Answer 22

Example:

Find the vertex,
y=x²+4x-3

y=x²+4x+4-4-3

Notice that (x²+4x+4) is a "perfect square trinomial" (you have to know what this term means).
y=(x²+4x+4)-7
y=(x+2)²-7

vertex is (2,7)
(and since parabola opens up the vertex is the minimum point.
the minimum is -7, and it occurs at x=2)

Answer 23

I made an error

Answer 24

4p is same as `a` in the other equation
if this is not conics section then i guess you should expand it first

Answer 25

the vertex in my example should be (-2,-7)

Answer 26

unit*

Answer 27

i have not heard of conics. so i think that i will stick with the basic a(x - h)^2 + k

Answer 28

alright then you should expand it

Answer 29

so i got:
\[\frac{1}{2}y + 2 = x^2 - 14x + 49\]

Answer 30

\(\color{blue}{\text{Originally Posted by}}\) @calculusxy
\[\frac{1}{2}y + 2 = x^2 - 14x + 49\]
\(\color{blue}{\text{End of Quote}}\)
that's correct solve for y

Answer 31

1/2y + 2 = x^2 - 14x + 49
1/2y = x^2 - 14x + 47
y = 2x^2 - 28x + 94

Answer 32

now two ways to find the vertex
you can complete the square or use -b/2a formula
up to you :=))

Answer 33

-(-28)/4 = 7
2(49) - 28(7) + 94
98 - 196 + 94
-4

(7, -4)

Answer 34

2(x - 7)^2 - 4

Answer 35

\(\huge\color{green}{\checkmark}\)

Answer 36

thanks!