The vertex of a parabola is at (12,-4) and its axis of symmetry is vertical. One of the x-intercepts is at (18,0). What is the x-coordinate of the other x-intercept?
+37159 The other x-axis intercept will be the same distance from the vertex as the given intercept..EXCEPT it will be on the OTHER SIDE of the axis of symmetry .
(6,0)
+876 Equation of a parabola is $y = a(x-h)^2 + k$
The vertex is $(h,k)$
We plug in and get $y = a(x-12)^2 - 4$
Plug in the coordinates: $0 = a(6^2) - 4$
$0 = 36a - 4$
$a = \frac{1}{9}$
The equation is now $y = \frac{1}{9}(x - 12)^2 - 4$
The other x-intercept is when $y=0,$ so we have $4 = \frac{1}{9}(x-12)^2$
$36 = x^2 - 24x + 144$
$x^2 - 24x + 108 = 0$
$r_1 + r_2 = 24$
$r_1 r_2 = 108$
We know that one root is $18,$ so we have $18 + r_2 = 24$ giving the answer to be $(24,0).$ The x-coordinate of that is $24.$
+37159 The other x-axis intercept will be the same distance from the vertex as the given intercept..EXCEPT it will be on the OTHER SIDE of the axis of symmetry .
(6,0)
