Find a quadratic function in standard form for each set of points
(0,3)(1,-4)(2,-9)
a. f(x)=x^2-27x+3
b. f(x)=x^2+8x-3
c.f(x)=x^2-8x+3
d. f(x)=x^2+27x+3

Question

Find a quadratic function in standard form for each set of points 
(0,3)(1,-4)(2,-9)
a. f(x)=x^2-27x+3
b. f(x)=x^2+8x-3
c.f(x)=x^2-8x+3
d. f(x)=x^2+27x+3

big_mike_lifting · Answer

Alright, basically to find the correct answer you can do trial and error, by taking each possible answer, and plugging in the points. Make sure you try all three, because the first two may work, and not the third.

laquino5220 · Answer

Thank you!

sshayer · Answer

$$y=ax^2+bx+c$$
find a,b,c by the given condition
i.e., all three points lie on it.

whpalmer4 · Answer

I will do an example with a different set of points.

$$(0,1),(1,3),(2,4)$$

We set up the equation $$y=ax^2+bx+c$$with each set of points, giving us 3 equations in 3 unknowns ($a,b,c$):

point $(0,1)$:
$$1 = a(0)^2 + b(0) + c$$$$1 = c$$that was easy!  Always nice when $x=0$ is one of the known points.
point $(1,3)$:
$$3 = a(1)^2 + b(1) + c$$$$3 = a+b+1$$

point $(2,4)$:
$$4 = a(2)^2+b(2)+c$$$$4 = 4a + 2b + 1$$

now we use our latter two equations (the only ones containing $a,b$) to find the values of $a,b$:

$$3=a+b+1$$$$4=4a+2b+1$$
we can rearrange the first to $$2-b=a$$and substitute $(2-b)$ wherever we see $a$ in the other equation:
$$4 = 4(2-b) + 2b + 1$$$$4 = 8-4b+2b+1$$$$-5=-2b$$$$b=\frac{5}{2}$$

Now plug that value into our substitution equation to find the value of $a$:
$$a = 2-b = 2 - \frac{5}{2} = -\frac{1}{2}$$

So our equation for a quadratic that fits those points is
$$y = -\frac{1}{2}x^2+\frac{5}{2}x+1$$

and if you plug in the values we are given, you will see that the equation produces them all exactly.

whpalmer4 · Answer

Note that @Big_Mike_Lifting's approach will find you the answer, but only because the answer has been provided.  You'd be stuck if you had to actually find the answer and only learned his approach...