Parabola: finding the a and q values

• May 3rd 2010, 09:56 AM
Mixxie16
Parabola: finding the a and q values
I need help with finding the a and q values if the defining equation f is f(x)= ax^2 + q

with the x and y values being (-3;6) and (3;6)
x y x y

How would u go about solving this equation :)
• May 3rd 2010, 11:25 AM
shenanigans87
Quote:

Originally Posted by Mixxie16
I need help with finding the a and q values if the defining equation f is f(x)= ax^2 + q

with the x and y values being (-3;6) and (3;6)
x y x y

How would u go about solving this equation :)

Notice what happens when x is 3 or -3

$3^2=9$.

But we know that the y value is supposed to equal 6, not 9. So the q value should subtract $9-6=3$.

$y=x^2-3$ suits the two ordered pairs you were given.
• May 3rd 2010, 01:22 PM
Quote:

Originally Posted by Mixxie16
I need help with finding the a and q values if the defining equation f is f(x)= ax^2 + q

with the x and y values being (-3;6) and (3;6)
x y x y

How would u go about solving this equation :)

Hi Mixxie,

if a=1, then q=-3

You need 2 clues to find "a" and "q".

The two pieces of information you are given are unfortunately
not giving two simultaneous equations as they are both

$9a+q=6$

since $(-3)^2=3^2$

The graph of the function is parabolic, the axis of symmetry is the y-axis.

$f(0)=q$

There are an infinite number of solutions until another clue is given.

$a=2,\ 2(9)+q=6,\ q=-12$

$a=3,\ 3(9)+q=6,\ q=-21$

etc