# 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

\$\displaystyle 3^2=9\$.

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

\$\displaystyle 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

\$\displaystyle 9a+q=6\$

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

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

\$\displaystyle f(0)=q\$

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

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

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

etc