Question

use the information provided to write the vertex form equation of each parabola.

1. y = -7x^2 - 14x - 6

Answer 1

factor the first two terms out of -7, please.

Answer 2

Note that:
\(\color{#000000 }{ \displaystyle -7x^2=(-7)\cdot x^2 }\) and,
\(\color{#000000 }{ \displaystyle -14x=(-7)\cdot 2x }\)

Answer 3

can you factor \(\color{#000000 }{ \displaystyle -7x^2-14x }\) out of -7?

Answer 4

trying

Answer 5

sure ...

Answer 6

@SolomonZelman hm, im confused

Answer 7

\(\color{#000000 }{ \displaystyle ab+ac=a\cdot (b+c) }\)

did you know?

Answer 8

Yeah I've seen the formula before but never really knew how to use it

Answer 9

Well, think about

2•4 + 8•4 = 4•(8+2) (factored out of 4)

and in fact if you work both sides,
8 + 32 = 4 • (10)
40 = 40

Answer 10

So if you see two components that are both products, and both have some common term "a", then this "a" can be factored out like I did with "a" in the formla, and with "4" in the example.

Answer 11

\(\color{#000000 }{ \displaystyle -7x^2 - 14x}\)

So wouldn't it make sense to say;
\(\color{#000000 }{ \displaystyle (-7)\cdot x^2 + (-7)\cdot 2x}\)

\(\color{#000000 }{ \displaystyle (-7)(x^2 +2x)}\)

Answer 12

ohhh okay i see now

Answer 13

So,
\(\color{#000000 }{ \displaystyle y = -7x^2 - 14x - 6}\)

\(\color{#000000 }{ \displaystyle y=(-7)(x^2 +2x)-6}\)

Answer 14

Ok, and do you know that:

\(\color{#000000 }{ \displaystyle (x+a)^2=x^2+2ax+a^2}\) ?

Answer 15

hmm, no haven't seen that formula before

Answer 16

\(\color{#000000 }{ \displaystyle (x+a)^2=}\)
\(\color{#000000 }{ \displaystyle (x+a) \cdot (x+a)=}\)
\(\color{#000000 }{ \displaystyle x\cdot (x+a)+a\cdot (x+a)=}\)
\(\color{#000000 }{ \displaystyle x\cdot x+x\cdot a+a\cdot x+a\cdot a=}\)
\(\color{#000000 }{ \displaystyle x^2+xa+ax+a^2=}\)
(xa is same as ax) \(\color{#000000 }{ \displaystyle x^2+2ax+a^2.}\)

Answer 17

I am doing a quick derivation of that...

Answer 18

you can look more into detail about
\(\color{#000000 }{ \displaystyle x^2+2ax+a^2=(x+a)^2}\)

but for know, you need to get
\(x^2+2x\) into that form.

Answer 19

you can see that in our case "a" is?
(Hint: 2ax in the formula coressponds to 2x in our case)

Answer 20

a is -7

Answer 21

\(\color{#000000 }{ \displaystyle x^2+2ax+a^2=(x+a)^2}\)
\(x^2+2x\)

2x=2ax
a=1

Answer 22

1²=1

Answer 23

So you need to add 1 to x²+2x to make it into the form of
\(\color{#000000 }{ \displaystyle x^2+2ax+a^2~~~~~~~~~~~=(x+a)^2}\)

Answer 24

but we can't just add 1, because that changes the value however we can;

\(\color{#000000 }{ \displaystyle y=(-7)(x^2 +2x)-6}\)

\(\color{#000000 }{ \displaystyle y=(-7)(x^2 +2x+1-1)-6}\)

The rule is:
\(\color{#000000 }{ \displaystyle (A)(C-D)=AC-AD}\)

(you can show this, you play with
5(5-3) = 5•5 - 5•3)

Answer 25

\(\color{#000000 }{ \displaystyle y=(-7)(x^2 +2x+1-1)-6}\)

in our case
\(\color{#000000 }{ \displaystyle x^2 +2x+1}\) is C
and -1 is -D. (or [the last] 1 is D)

Answer 26

\(\color{#000000 }{ \displaystyle y=(-7)(x^2 +2x+1)-(-7)(1)-6}\)
\(\color{#000000 }{ \displaystyle y=(-7)(x^2 +2x+1)+7-6}\)
\(\color{#000000 }{ \displaystyle y=(-7)(x^2 +2x+1)+1}\)

Answer 27

notice that
(x+1)²=x^2+2(1)(x)+1² = x² + 2x +1

so re-write the part in parenthesis.

Answer 28

and then the vertex of
\(\color{#000000 }{ \displaystyle y=a(x-h)^2+k}\)
is (h,k)

and equivalent the vertex of
\(\color{#000000 }{ \displaystyle y=a(x+t)^2+k}\)
is (-t,k)

Answer 29

oh man thank you dude, i'm gonna have to read through all this again, i just can't get it

Answer 30

yes, and ask other people questions about anything you don't understand ... to ask is most important.

Answer 31

Good luck

Answer 32

Here's a slightly different approach:
Your equation is a parabola:
\[y = -7x^2 - 14x - 6\]

If we have a parabola written in the form \[y = ax^2 + bx + c\]where \((a,b,c)\) are all constants as they are here (\(a=-7,\ b=-14,\ c=-6)\) we can find the vertex easily by knowing that the \(x\)-value of the vertex (which we will call \(h\)) will be\[x = -\frac{b}{2a}\]and the \(y\)-value (which we will call \(k\)) we get by just plugging the value of \(x\) into the formula of the parabola.

Once you know the vertex is at \((h,k)\) you can write the formula for the parabola in vertex form using this:
\[y = a(x-h)^2 + k\]
The value of \(a\) is the same value we had for \(a\) when we found the vertex.