Question

what are the vertex and axis of symmetry of the equation y=-2x^2+8x-18

Answer 1

https://www.mathway.com/

Answer 2

i tried it dont help

Answer 3

oh ok

Answer 4

would you help me i dont know how to find it

Answer 5

give me a minute

Answer 6

vertex is the high point, yeah?

Answer 7

look it up on google n the answer is there it was already asked on openstudy hope that helped.

Answer 8

i am with you jack

Answer 9

y=−2x2+8x−18

y+18−8=−2(x2−4x+4)

y+10=−2(x−2)2

Answer 10

cool, so ur looking for ur turning point
ur equation's:
y=-2x^2+8x-18

Answer 11

yes

Answer 12

y = -2x^2+8x-18

so ur derivative's (y')
y' =-4x + 8

Answer 13

you've done derivatives yet, yeah?

Answer 14

no not yet

Answer 15

i am with you though

Answer 16

i have answer choices would that help you in away

Answer 17

dam...ok the short of derivatives is:
bring the power down, multiply by co-efficient, then put back up and minus 1

so if u had:
y = 10x^4 -12x^3 +9x^2 + 2x + 19

derivative would be
y' = 10*4 x^4-1 - 12*3 x^3-1 + 9*2 x*2-1 + 2*1 x^1-1 + 0

so
y' = 40x^3 - 36x^2 + 18x*1 + 2x^0
y' = 40x^3 - 36x^2 + 18x*1 + 2

make sense?

Answer 18

either way: you're derivative is the slope of the curve at any x or y point along it

so
y' =-4x + 8

when slope = 0, its a turning point

so
y' =-4x + 8
0 =-4x + 8
-8 =-4x
x = 2

your vertex occurs at x = 2

whats the y value at this point? @jamesx3rd ?

Answer 19

yeah you times it the minus the power down by one

Answer 20

i see how you get the answers i am with i am actualy learning this from you

Answer 21

what do you do with the x=2 now though

Answer 22

were did the y go?

Answer 23

hall good dude, happy ur learning

y = -2x^2+8x-18
if x = 2

y = -2x^2+8x-18
y = -2(2^2) +8(2) -18
y = -2*4 + 8*2 - 18
y = -8 + 16 - 18
y = -10

so when x = 2, y = -10

so vertex occurs at (2, -10)

Answer 24

my answer choices are:
Vertex: (2,-10) axis of symmetry x=2 i think this one is right
Vertex: (2,-10) axis of symmetry x =-10
Vertex: (-2,-10) axis of symmetry x=-2
Vertex (-2,10) axis of symmetry x=-2

Answer 25

you plug the 2 in the problem and got that -10 right

Answer 26

may be helpful in the meantime, until u get onto derivatives in a big way :

The standard equation of a parabola is
y = ax2 + bx + c.

But the equation for a parabola can also be written in "vertex form":
y = a(x – h)2 + k

In this equation, the vertex of the parabola is the point (h, k).

The coefficient of x here is –2ah. This means that in the standard form, y = ax2 + bx + c, the expression

-b/2a gives the x-coordinate of the vertex.

Answer 27

so was the first one right

Answer 28

sauce is http://hotmath.com/hotmath_help/topics/vertex-of-a-parabola.html

Answer 29

and yep, a is correct

Answer 30

would you help me on one more problem ill make another post

Answer 31

yep ill follow

Answer 32

thanks