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.#

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#