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# help parabola

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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?

Jun 30, 2021

#2
+36431
+2

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)

Jun 30, 2021

#1
+874
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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 $24.$

Jun 30, 2021
#2
+36431
+2

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)

ElectricPavlov Jun 30, 2021