Find the vertex and the x-intercepts (if any) of the parabola. (If an answer does not exist, enter DNE.)
f(x) = -16x2 + 4x + 2
To find the vertex and the x-intercepts of a parabola represented by a quadratic function, we'll follow these steps:
Step 1: Identify the coefficients in the given quadratic function.
In the function f(x) = -16x^2 + 4x + 2, the coefficients are:
a = -16 (coefficient of x^2)
b = 4 (coefficient of x)
c = 2 (constant term)
Step 2: Find the x-coordinate of the vertex using the formula -b/(2a).
Using the formula, we can substitute the values of a and b into the formula to find the x-coordinate of the vertex.
x = -b / (2a)
x = -4 / (2 * -16)
x = -4 / -32
x = 1 / 8
So, the x-coordinate of the vertex is 1/8.
Step 3: Find the y-coordinate of the vertex by substituting the x-coordinate into the quadratic function.
To find the y-coordinate of the vertex, we substitute the x-coordinate (1/8) back into the original function.
f(1/8) = -16(1/8)^2 + 4(1/8) + 2
f(1/8) = -16/64 + 4/8 + 2
f(1/8) = -1/4 + 1/2 + 2
f(1/8) = -1/4 + 2/4 + 8/4
f(1/8) = 9/4
Therefore, the y-coordinate of the vertex is 9/4.
The vertex of the parabola is (1/8, 9/4).
Step 4: Find the x-intercepts (if any) by solving the quadratic equation for f(x) = 0.
To find the x-intercepts, we set the quadratic equation equal to zero and solve for x. In other words, we solve -16x^2 + 4x + 2 = 0.
Using the quadratic formula x = (-b ± √(b^2 - 4ac)) / (2a), we can plug in the values a = -16, b = 4, and c = 2.
x = (-4 ± √(4^2 - 4(-16)(2))) / (2 * -16)
x = (-4 ± √(16 + 128)) / -32
x = (-4 ± √144) / -32
x = (-4 ± 12) / -32
Simplifying further, we have:
x = (8) / -32 or x = (-16) / -32
x = -1/4 or x = 1/2
Therefore, the x-intercepts of the parabola are -1/4 and 1/2.
In summary, for the given parabola f(x) = -16x^2 + 4x + 2, the vertex is (1/8, 9/4), and the x-intercepts are -1/4 and 1/2.