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

2 Answers
Aug 29, 2017

3x^2-5x+2.3x25x+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.

Aug 29, 2017

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