Question

maximum or minimum point for (x-4)^2-5
And maximum or min value?

Answer 1

@ganeshie8

Answer 2

can a square of something be negative ?

Answer 3

(3)^2 = ?
(-3)^2 = ?

Answer 4

It's given as x^2-8x+11 then we have a to convert it to vertex form

Answer 5

Then it x-4^2-5

Answer 6

Nice :)
first of all notice that the square of something can never be negative, the minimum value is 0

Answer 7

so the minimum value of \(\large (x-4)^2 - 5 \) is \(\large 0 - 5\)

Answer 8

which equals \(\large -5\)

Answer 9

How is it 0?

Answer 10

And how do we know if it's min or max

Answer 11

+still I have to find the min or max point

Answer 12

@ganeshie8

Answer 13

if you prefer memorizing :
\[\large y = a(x-h)^2 + k\]
has a minimum point if \(\large a \) is positive,
has a maximum point if \(\large a \) is negative`

Answer 14

the point would be the vertex : \(\large (h, k)\)
and the value is the y coordinate \(\large k\)

Answer 15

\[\large y = 1(x-4)^2-5\]

compare it with the standard quadratic :
a = 1 which is positive, so it will have a minimum point

Answer 16

Ok

Answer 17

Than its minimum point is -4 and -5

Answer 18

And min value is -5

Answer 19

?

Answer 20

Is it right ?

Answer 21

And I wanna ask u ...... What is the diff between maximum point and maxium value

Answer 22

minimum point is (4, -5)
not (-4, -5)

Answer 23

thats a good question, do you know how the graph looks ?

Answer 24

Yes downward

Answer 25

it is a valley,
a valley has a minimum point at its bottom

Answer 26

notice that it has no maximum point because the graph increases forever in both ends, there is no end to how high it can go

Answer 27

I mean maximum point and maximum value

Answer 28

yes a downward facing graph has a maximum point, it looks like a hill :