Question

finding vertex in standard form?
y=-x^2+2x-5

Answer 1

I understand how to get the x coordinate but i keep getting 8 for the y coordinate, and the book says it is 1, 6. can someone help me find my error

Answer 2

\[\large \bf ax^2+bx+c=0\]

and discriminant,\(\large \bf D=b^2-4ac\)

to find vertex ,we use this formula :-
\[\large \bf Vertex(\frac{-b}{2a},\frac{-D}{4a})\]

Answer 3

can you now solve it ? @bastiennecroix

Answer 4

The method we learned for finding the y is just plugging the x back into the quadratic

Answer 5

Is it possible to find it that way? plugging it back in ? @mayankdevnani

Answer 6

yeah.Just plug all the values in formula and definitely,you will get correct answer.

Answer 7

so, (-1)^2 + 2(1) + 5 ?

Answer 8

sorry, the original quadratic is y= -x^2+2x+5

Answer 9

@mayankdevnani

Answer 10

i can tell you one answer and other answer is for you :)

Answer 11

see im confused because when that's added up, it's 8.

Answer 12

to find x-coordinate,
\[\large \bf \frac{-b}{2a}=\frac{-2}{2(-1)}=\frac{-2}{-2}=1\]

Answer 13

therefore,x-coordinate is 1

Answer 14

Yes. and so to find the y, we need to plug the x back into the quadratic

Answer 15

so: -(1)^2 + 2(1) +5

Answer 16

yeah !!

Answer 17

and that equals 8, unless my brain is messing up

Answer 18

So I was right? My textbook says the vertex is 1, 6

Answer 19

I've been going through this problem forever trying to see where I went wrong

Answer 20

oh!!!
\[\large \bf -(1)^2+2+5=-1+2+5=6\]

Answer 21

I thought squared negatives were always positive

Answer 22

no,the negative sign is out of the bracket,so +ve squared is positive

Answer 23

OH! thank you so much!!!!!!

Answer 24

welcome :)