what is the range of f(x)=2x^2-4x-3

Question

helpppppp meeee plzzzz what is the range of f(x)=2x^2-4x-3

Answer 1

f(x) = 2x² - 4x - 3

We know this is a parabola because the highest degree of x is 2. Seeking to convert this equation into vertex form:

y = 2x² - 4x - 3

(y + 3) / 2 = x² - 2x

1 + (y + 3) / 2 = x² - 2x + 1 this is the "completing the square" method

1 + (y + 3) / 2 = (x - 1)²

(y + 5) / 2 = (x - 1)²

y + 5 = 2(x - 1)²

y = 2(x - 1)² - 5

From vertex form y = a(x - h)² + k, we know that (h, k) is the vertex. So from y = 2(x - 1)² - 5, the vertex of our parabola is (1, -5). Since our original expression f(x) = 2x² - 4x - 3 has a positive sign in front of x², we know the parabola is oriented upwards. This means that the minimum value of the function is at the vertex, since y increases as x varies from 1 in the positive or negative direction. Therefore, the range of the function is y ≥ -5, or [-5, ∞).

Answer 2

The function f(x) = 2x²-4x-3 is a quadratic function, and its function is a parabola. Since the coefficient of x² , 2, is positive, the function values increase without bound as x increases without bound as x increases without bound or decreases without bound. (That is, the function values goes to (positive) infinity in both directions. In other words, the parabola opens upward.)

To determine the minimum value of this function, it would help to rewrite the formula for the function in vertex form. The y-coordinate of the vertex will be the minimum value.

To find the vertex form, recall that if y = ax²+bx+c, the vertex form is given by

y = a(x-h)²+k, where h = -b/(2a) and k = c - b²/(4a). In our case, a = 2, b = -4, and c = -3.

Thus, h = -(-4)/(2(2)) = 1, and k = -3 - (-4)²/(4(2)) = -5. Hence the vertex form is

f(x) = 2(x-1)²-5, and the vertex is the point (1,-5).

We can conclude that the range of f is the set {x | x ≥-5}, which can be written as [-5,∞) in interval notation.