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
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
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