Find the vertex of the functions given below.
1. y=(x-4)(x+2)
2. y=2x^2-4x+1
The midpoint of the x-intercepts of a parabola has the same x value as the vertex.
the x-intercepts are obviously x=4 and x=-2
so the x value of the vertex is......
sub that back into the equation to get the y value.
for the second one, just use the method of completing the square.
To find the vertex of each function, we will use different methods for each function.
1. Function: y = (x - 4)(x + 2)
To find the vertex, we first need to find the x-coordinate of the vertex using the midpoint of the x-intercepts method. The x-intercepts are the values of x where the function crosses or intersects the x-axis. In this case, we can see that the x-intercepts of the function are x = 4 and x = -2.
Now, to find the x-coordinate of the vertex, we can take the average of the x-intercepts:
x-vertex = (4 + (-2)) / 2 = 2 / 2 = 1
So, the x-coordinate of the vertex is 1.
Next, we substitute the x-coordinate of the vertex back into the equation to find the y-coordinate of the vertex. We substitute x = 1 into the equation and solve for y:
y = (1 - 4)(1 + 2) = (-3)(3) = -9
Therefore, the vertex of the function y = (x - 4)(x + 2) is (1, -9).
2. Function: y = 2x^2 - 4x + 1
To find the vertex using the method of completing the square, we need to rewrite the function in vertex form, which is expressed as y = a(x - h)^2 + k. The values h and k represent the x-coordinate and y-coordinate of the vertex, respectively.
First, let's complete the square to rewrite the function:
y = 2x^2 - 4x + 1
To complete the square, we need to add and subtract a constant term. In this case, we will add and subtract (4/2)^2 = 4 to the equation:
y = 2(x^2 - 2x + 1) - 4 + 1
Now, we can factor the perfect square trinomial (x^2 - 2x + 1) as (x - 1)^2:
y = 2(x - 1)^2 - 3
Comparing this equation to the vertex form, y = a(x - h)^2 + k, we can see that the vertex of the function is (h, k) = (1, -3).
Therefore, the vertex of the function y = 2x^2 - 4x + 1 is (1, -3).