Question

Find a formula for a quadratic function that passes through the points (2,0), (-1,0) and (0,4)

Answer 1

The general quadratic function: \(y=ax^2+bx+c\).

Plug in your points:
\[\begin{cases}4=0a+0b+c\\0=a-b+c\\0=4a+2b+c\end{cases}\]
Solve for \(a,b,c\).

Answer 2

So I plug in x and y from one of the points?

Answer 3

I believe that you know from the points that the x values are 2, -1, and 0.

And you can make your binomials (remember the sings change) (we just just leave the 0 one out because it wont make a difference on the answer)

(x-2)(x+1)(x+0)

Answer 4

Do you just foil it out with this method?

Answer 5

Yup!

Answer 6

I'm pretty sure that's what they want you to do

@sithsandgiggles correct me if I'm wrong

Answer 7

(x^2+x-2)(x+0)
x^3 + 0 + x^2 + 0 - 2x - 0

Answer 8

@iambatman

Answer 9

@Kainui

Answer 10

(x^2-x-2)(x+0)

-2+1 is -1

Answer 11

x^3+x^2-2x

Answer 12

x^3-x^2-2x

Answer 13

I am thinking that I have to solve three linear equations to find the formula for this quadratic. I'm not sure about your way @marissalovescats

Answer 14

@dan815

Answer 15

look at sithsandgigles first post

Answer 16

he's solved it for you

Answer 17

so I have to solve all three equations for a, b, and c?

Answer 18

all quadratic eqn can be put in the form y=ax^2+bx+c we have to find the a b and c constant that satisfy your points

Answer 19

yes

Answer 20

tell me if u get stuck on something else

Answer 21

thanks dan

Answer 22

1) 0 = 4a + 2b + c
2) 0 = a + -b + c
3) 4 = a + b + c

(0 = 4a + 2b + c)
0 = a - b + c

0 = -3a - b

Answer 23

can someone verify this is right?

Answer 24

@dan815 @satellite73

Answer 25

@mathslover

Answer 26

i think it is way easier than all of that

Answer 27

\[ (2,0), (-1,0) \] tells you the zeros are \(2\) and \(-1\) so it must factor as
\[f(x)=a(x-2)(x+1)\]

Answer 28

since \(f(0)=4\) you know
\[a(0-2)(0+1)=-2a=4\] making \(a=-2\)

Answer 29

your quadratic is therefore
\[f(x)=-2(x-2)(x+1)\]

Answer 30

I thought there was an easier way, I was doing about 20 steps with solving both 3 equations and then substituting two of the equations to eliminate c and then substituting the bottom two equations

Answer 31

you would have to do that if you were not told the two zeros, but since you were, it was much easier than all that mess

Answer 32

how do we know it must factor like you did? @satellite73

Answer 33

we are not given the equation in ax^2+bx+c form

Answer 34

@marissalovescats, that method also works, but \((0,4)\) is not a root.