MHB Quadratic Equation Help: Find 3 Points to Determine Answer

[Monoxdifly]

In a graph , straight line intersects the parabola at(-3,9) & (1, 1) Then the equation is
A) x^2-2x+3=0
B) x^2+2x-3=0
C) x^2-3x+2=0
D) x^2-2x-3=0

I know that I can find the answer by substituting the known values to each options, but how to do it the proper way? We need at least three known points, right? How do we know where is the other point needed to solve this?

 

[MarkFL]

You are correct in that it takes 3 points to exactly determine a quadratic curve, however all of the options are of the form:

$$f(x)=x^2+ax+b$$

We only have 2 parameters to determine. So, we may write, using the 2 given points:

$$9-3a+b=9\implies 3a-b=0$$

$$1+a+b=1\implies a+b=0$$

This implies $$a=b=0$$, and so the only quadratic of the above two-parameter family passing through the given points is:

$$f(x)=x^2$$

 

[Monoxdifly]

So, the answer wasn't in the option at all?

 

[MarkFL]

So, the answer wasn't in the option at all?

It appears that's the case. :)

 

[Monoxdifly]

Okay, thanks.

 

[RLBrown]

In a graph , straight line intersects the SOLUTION SET at(-3,9) & (1, 1) Then the equation is
A) x^2-2x+3=0
B) x^2+2x-3=0
C) x^2-3x+2=0
D) x^2-2x-3=0
?

Answer C
(The graph of the solution set on (x,y) is two vertical lines.)

 

[skeeter]

In a graph , straight line intersects the parabola at(-3,9) & (1, 1) Then the equation is
A) x^2-2x+3=0
B) x^2+2x-3=0
C) x^2-3x+2=0
D) x^2-2x-3=0

the equation is where ...

an unknown quadratic function = an unknown linear function $\implies$ unknown quadratic function - unknown linear function = 0

in other words, the left side of each equation in the given choices is the simplified form of unknown quadratic function - unknown linear function

(A) $x^2-2x+3=0$, note the discriminant $b^2-4ac < 0 \implies$ no real solutions

(B) $x^2+2x-3=0 \implies (x+3)(x-1)=0 \implies x=-3 \text{ and } x=1$ are solutions

(C) $x^2-3x+2=0 \implies (x-2)(x-1)=0 \implies x=2 \text{ and } x=1$ are solutions

(D) $x^2-2x-3=0 \implies (x-3)(x+1)=0 \implies x=3 \text{ and } x=-1$ are solutions

... which set of x-value solutions match up with the abscissas for the given points of intersection?Poorly worded question, imho.