A graph of a parabola has a line of symmetry X=3 and contains the points (1,0) and (4,-3) Determine an equation for this parabola.

using y=a(X-h)^2+K

-3 = a(4-1)^2 + 0

-3 = -1/3(4-1)^2 + 0

did i work this correctly?

thanks

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- Jan 25th 2009, 07:30 PMSåxonDetermine an equation for this parabola
A graph of a parabola has a line of symmetry X=3 and contains the points (1,0) and (4,-3) Determine an equation for this parabola.

using y=a(X-h)^2+K

-3 = a(4-1)^2 + 0

-3 = -1/3(4-1)^2 + 0

did i work this correctly?

thanks - Jan 25th 2009, 07:48 PMJhevon
um, no

our parabola is of the form $\displaystyle y = a(x - 3)^2 + k$

since $\displaystyle (1,0)$ is on the graph, we have

$\displaystyle 0 = a(1 - 3)^2 + k$ .................(1)

since $\displaystyle (4,-3)$ is on the graph, we have

$\displaystyle -3 = a(4 - 3)^2 + k$ ................(2)

thus we have two simultaneous equations with two unknowns. solve this system to find your parabola