A parabola with a vertex at (–1, –2) passes through the point (1,6). Using this information,
what are the x-intercepts of this parabola?
A) (1 – , 0) and (1 + , 0)
B) (2, 0)
C) (0, 0) and (–2, 0)
D) (0, 0)
how would i figure this out?
Symmetric about vertical line x =-1
so 0 and -2 are the only x values that are the same distance less than or more than -1
45
a
To find the x-intercepts of a parabola, you need to find the values of x where the parabola intersects the x-axis. In other words, these are the points on the parabola where y = 0.
Given that the parabola has a vertex at (-1, -2) and passes through the point (1,6), we can assume that the parabola is in the form y = a(x - h)^2 + k, where (h, k) represents the vertex coordinates.
Let's substitute the vertex coordinates into the equation to find the value of a:
-2 = a(-1 - (-1))^2 + (-2)
-2 = a(0)^2 + (-2)
-2 = a(0) - 2
-2 = -2
This tells us that a = 1. So we now have the equation for the parabola as y = (x - (-1))^2 + (-2).
To find the x-intercepts, we set y = 0 and solve for x:
0 = (x - (-1))^2 + (-2)
0 = (x + 1)^2 - 2
Now, let's solve this equation:
(x + 1)^2 - 2 = 0
(x + 1)^2 = 2
To simplify, let's take the square root of both sides:
√((x + 1)^2) = √2
x + 1 = ±√2
Now, isolate x:
x = -1 ± √2
Therefore, the x-intercepts of the parabola are (-1 - √2, 0) and (-1 + √2, 0).
So, the correct answer is A) (-1 - √2, 0) and (-1 + √2, 0).