Question

How do I find the vertex of a trinomial if there are no x-intercepts? How do I find the vertex of a trinomial if there are no x-intercepts? @Mathematics

Answer 1

-b/(2a)

Answer 2

x = -b/(2a) no matter what the trinomial is

Answer 3

and then, after you find x, insert it into the trinomial to find y

Answer 4

If you don't understand how to do that, then you should just drop out of math forever

Answer 5

NotFlunk

Answer 6

waazaat yzw tkn bout?

Answer 7

why?

Answer 8

x^2-2x+1 has no x-intercepts on the y axis,

but it has a vertex of

x = -b/2a = -(-2)/2(1) = 2/2 = 1

y = (1)^2-2(1)+1 = 1-2+1 = 2-2 = 0

so it has a vertex of (1,0)

Answer 9

That's an example

Answer 10

For\[3x ^{2}-9x+16, I got 9, 178.\]

Answer 11

That's definietyle rong.

Answer 12

3x^2-9x+16

x = -b/(2a) = 9/9 = 1
y = 3(1)^2-9(1) + 16 = 3-9+16 = 19-9 = 10

so the vertex of 3x^2-9x+16 = (1,10)

Answer 13

Dude, that wasn't even close man. How did you get (9,178) ?

Answer 14

No according to hmy graph though...The vertex is at 1.5, 9.5 (somehwere, noexact position)

Answer 15

Well, you must have graphed it wrong somewhere

Answer 16

unless you did a rough est kidna calculation.

Answer 17

NO, I did an exact calculation. You can't roughly calculate a vertex

Answer 18

However, you calculator can roughly estimate the graph of a function.

Answer 19

I don't think you found the exact vertex of your graph man

Answer 20

noes. err, was your a 3 and your b 9?

Answer 21

-9

Answer 22

Oh I found my mistake. Let me re-cal

Answer 23

its 1.5,9.25 right?

Answer 24

x = -b/(2a) = -(-9)/2(3) = 9/6 = 3/2


Yeah but the sad thing is, you still didn't use the formula to find it.

Answer 25

You had to graph it

Answer 26

I feel sorry for you man

Answer 27

itws part of a larger assignment.

Answer 28

something with calcs and graphs.

Answer 29

thx anyways.