Question

How do you write a quadratic function in standard form whose graph passes through points (1,0), (2,4), (0,2)?

Answer 1

`3x^2-5x+2.`

Explanation:

Suppose that, the reqd. quadr. fun. is,

`y=f(x)=ax^2+bx+c, (x in R)(a,b,c in RR, a!=0)......(star).`

`f" passes through "(1,0) rArr (x,y)=(1,0)" must satisfy "(star).`

`:. 0=a*1^2+b*1+c, i.e., a+b+c=0.......(star^1).`

Similarly, `(2,4) in f rArr 4a+2b+c=4..........(star^2),` and,

`(0,2) in f rArr c=2..........................................(star^3).`

`(star^3) and (star^1) rArr a+b+2=0, or, b=-a-2,` and,

sub.ing this in `(star^2), 4a+2(-a-2)+2=4 rArr a=3,` &, so,

`b=-a-2=-3-2=-5.`

Thus, `a=3, b=-5, c=2,` give us the desired quadr. fun.

`f(x)=3x^2-5x+2.`

Answer 2

`3x^2-5x+2=0`

Explanation:

Since you are given three points, three equations can be created, all in the form of `ax^2+bx+c=y`, which is used for quadratic functions.

So, the first equation using point `(1,0)` is: `a(1)^2+b(1)+c=0`
Which simplifies as `a+b+c=0`
The second equation using `(2,4)` is `4a+2b+c=4`
The third equation with `(0,2)` is `c=2` since `x=0` so the "a" and "b" parts become 0 as well.

Now that you have three equations, you can solve for the three variables using substitution and elimination.

`a+b+c=0`
`4a+2b+c=4`
`c=2`

Multiplying the first equation by 2, you get `2a+2b+2c=0`
Subtracting this equation from `4a+2b+c=4`, this is obtained: `2a-c=4`
You can now substitute in your third equation, `c=2` into `2a-c=4`
You then get `2a-2=4` which simplifies to `2a=6` so
`a=3`

Now that you know what "a" and "c" are, you can plug this into the second equation.
So, `12+2b+2=4`
`b=-5`

One more step!
Your final equation is in the form of `ax^2+bx+c=0`
Just put all the values into the corresponding places, and you get:
`3x^2-5x+2=0`