Question

Quadratic formula!

Answer 1

I'm not sure what the vertex would be

Answer 2

I have 40 minutes to do 19 questions lol cause they are taking me forever to understand

Answer 3

Easy way to find vertex when \[y=ax^2+bx+c\]is the x coordinate is at \[x = -\frac{b}{2a}\]
Plug in the value of \(x\) to find the corresponding value of \(y\)

Answer 4

f(x)=x^2-8x+7
f'(x)=2x-8
when f'(x)=0 then its the vertix point ( for Quadratic equations)
2x-8=0, x= 4
so vertex= (4,-9)

Answer 5

so how do i know the y and x

Answer 6

You have a formula
\[y = x^2-8x+7\]Compare with \[y=ax^2+bx+c\]\[a = 1, b = -8, c = 7\]Right?

Answer 7

Is x (7,0)
and y (0,1)

Answer 8

hmm @whpalmer4 why ur formula got -ve sighn :o

Answer 9

Vertex x-coordinate is at \[x = -\frac{b}{2a} = -\frac{-8}{2*1} = 4\]\[y = (4)^2-8(4)+7 = 16 - 32 + 7 = -16 +7 = -9\]
So vertex is at \((4,-9)\)

Answer 10

@ikram002p

if \[y = ax^2+bx+c\]\[y' = 2ax + b\]\[2ax+b = 0\]\[2ax = -b\]\[x = -\frac{b}{2a}\]

Answer 11

ahh ok lol
its the same way i did it hehe

Answer 12

Guys.. is my x (7,0) and y(0,1)

Answer 13

well, except it's well known to any analytic geometry student who pays attention even if they don't know calculus :-)

No, look at what we did!

Answer 14

no , @vshiroky its wrong

Answer 15

I thought the vertex was different

Answer 16

so y is 0,-9

Answer 17

look at the sketck
the vertex is a point (x,y)=(4,-9)