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#### miller1991

##### New member

- Jan 23, 2019

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The Role of H in the quadratic function ( vertex form)

i get that this is how its written on a graph y=(x-2)^2+k

that the graph looks as if the value of h is positive as in +2 ( however its value is actually negative)

looks like it shifted right

my textbook contradicts itself

y=3(x-1)^2 +2 or y=3(x-(-1))^2 +2

a positive ie x minus a negative ie -1 equals a positive ( thus h is positive and shifts right

the value of h is +1 thus there is a right shift 1 unit

( two positives make a negative, or subtracting a negative from a positive gives a positive )

for the example

y=-1(x-3)^2

i need to know if the value of a affects the value of h

y=-1(x-(-3))^2

if a is -1

we looking at

(x-(-3))^2 gives a positive h value

( two positives make a negative, or subtracting a negative from a positive gives a positive )

then its basically

-1(3) equals a negative ( as in a negative times a positive equals a negative) thus switches h to a negative vaule

h is -3 the function shifts left

and on the graph reads as -1(x+3)^2 or w.e (not sure if that part is right)

in short a good place for me to start is understanding if A in vertex form affects the value of H in regards to the shift of h

i just need to know how to solve for the value of h

like

if the question is describe the shifting of h in this function

what does h do

and does a affect the value of h

textbook question

state the value of h and describe the shifting of the function

y=3(x-1)^2 +2

answer given H = 1 and shifts right 1 unit

y=-(x-3)^2

answer given

h= -3 and shifts left 3 units

as far as this book indicates a is affecting the h movement

THANKS GUYS !!!!!!! FOR HELPING ME GET further with this

w.e info you have to help me move forward would be appreciated.

so is it possible that a does not affect the shifting of h

and these

(x-(-3))^2

invisible brackets actually mean multiplication and i am not adding and subtracting here to find the value of h

i get that this is how its written on a graph y=(x-2)^2+k

that the graph looks as if the value of h is positive as in +2 ( however its value is actually negative)

looks like it shifted right

my textbook contradicts itself

y=3(x-1)^2 +2 or y=3(x-(-1))^2 +2

a positive ie x minus a negative ie -1 equals a positive ( thus h is positive and shifts right

the value of h is +1 thus there is a right shift 1 unit

( two positives make a negative, or subtracting a negative from a positive gives a positive )

for the example

y=-1(x-3)^2

i need to know if the value of a affects the value of h

y=-1(x-(-3))^2

if a is -1

we looking at

(x-(-3))^2 gives a positive h value

( two positives make a negative, or subtracting a negative from a positive gives a positive )

then its basically

-1(3) equals a negative ( as in a negative times a positive equals a negative) thus switches h to a negative vaule

h is -3 the function shifts left

and on the graph reads as -1(x+3)^2 or w.e (not sure if that part is right)

in short a good place for me to start is understanding if A in vertex form affects the value of H in regards to the shift of h

i just need to know how to solve for the value of h

like

if the question is describe the shifting of h in this function

what does h do

and does a affect the value of h

textbook question

state the value of h and describe the shifting of the function

y=3(x-1)^2 +2

answer given H = 1 and shifts right 1 unit

y=-(x-3)^2

answer given

h= -3 and shifts left 3 units

as far as this book indicates a is affecting the h movement

THANKS GUYS !!!!!!! FOR HELPING ME GET further with this

w.e info you have to help me move forward would be appreciated.

so is it possible that a does not affect the shifting of h

and these

(x-(-3))^2

invisible brackets actually mean multiplication and i am not adding and subtracting here to find the value of h

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