Question

Fill in the missing term so that the quadratic equation has a graph that opens up, with a vertex of
(- 3, - 25), and x intercepts at x = -8 and x = 2. (Do not include the negative sign in your answer.)

Answer 1

@matricked

Answer 2

assume
y = ax^2+bx+c
given
y=-25 when x=-3 and
y=0 when x= -8
y=0 when x=2

Answer 3

mmhhmmm?

Answer 4

Alternately
y =k (x+8)(x-2)
you just need to find k using (- 3, - 25),

Answer 5

how do i do that?

Answer 6

Find k like this, using (-3, -25)...

Answer 7

yes?

Answer 8

i want all the help i can get ;L

Answer 9

In y = k(x+8)(x-2), your x values will be -3, and your y value will be -25 (from your point (-3, -25)) Fill those in and solve for k:

Answer 10

i got -25=k(5-5) ;P

Answer 11

-25=k(-3+8)(-3-2)
-25=k(5)(-5)
-25=-25k
k=1

Answer 12

oh

Answer 13

so its...

Answer 14

You had the right idea above, you just put the two 5s in the same set of parenthesis instead of in their own andd multiplying them.

Answer 15

so its -1?

Answer 16

When you find k, or a as I like to call it, you fill it into your standard equation\[ax ^{2}+bx+c\]like this:

Answer 17

yes?

Answer 18

Use the roots as your factors. We have roots of -8 and 2. So the roots are (x + 8)(x - 2), right? Do this, then, with the k (or a) value:

Answer 19

ok

Answer 20

a(x+8)(x-2)-->1((x+8)(x-2)-->x^2+6x-16

Answer 21

yes

Answer 22

That's your equation in standard form...\[y=x ^{2}+6x-16\]

Answer 23

ohhh

Answer 24

thx man ;3

Answer 25

The clue is, find your "a" or "k" (same thing) by using your vertex and your roots. Then use the roots and multiply them by your a value to expand.

Answer 26

now if you could help me with a quadratic equation i made, and tell me if you can solve it using the factorisation, that would be really helpful ;3

Answer 27

ohh

Answer 28

And BTW...I'm not a man, I'm a ma'am. But that's ok!

Answer 29

2a^2+5x+14

Answer 30

oh sry ma'am

Answer 31

anyways that was the equation

Answer 32

you can't factor that because you have both an a and an x in it. Which is it? x or a?

Answer 33

oh sry

Answer 34

2x^2+5x+14

Answer 35

is that factorable?

Answer 36

With the quadratic formula, anything is factorable!!!

Answer 37

i need an equation that can't be factorised ;P

Answer 38

care to give me an example?

Answer 39

Ok, then, that can't be factored. The solutions when you use the quadratic formula are\[\frac{ -5\pm i \sqrt{87} }{ 4 }\]That is a polynomial with nonreal complex solutions. I'd say it would be fair to say that doesn't factor. Not at least the "regular" way! You can use it!

Answer 40

yay!

Answer 41

thank you!

Answer 42

Yay is right!

Answer 43

huzzah!

Answer 44

You're welcome!