Question

The vertex of a parabola is at (6,5), and its axis of symmetry is vertical. One of the x-intercepts is at (8, 0). What is the x-coordinate of the other x-intercept?

Answer 1

Answer:

$x=4$

Step-by-step explanation:

Vertex form of the quadratic equation: $f(x) = a(x - h)^2+k$

where the vertex is $(h, k)$

Given vertex = (6, 5):

$\implies f(x) = a(x - 6)^2+5$

If one the x-intercepts is (8, 0):

$\implies f(8) =0$

$\implies a(8 - 6)^2+5=0$

$\implies 4a+5=0$

$\implies a=-\dfrac{5}{4}$

Therefore, equation of parabola:

$\implies f(x) = -\dfrac54(x - 6)^2+5$

x-intercepts occur when f(x) = 0:

$\implies f(x)=0$

$\implies -\dfrac54(x - 6)^2+5=0$

$\implies -\dfrac54(x - 6)^2=-5$

$\implies (x - 6)^2=4$

$\implies x-6=\pm 2$

$\implies x=8, x=4$

Therefore, the x-coordinate of the other x-intercept is 4.