The function f(x) = (x-2)(x+4) is a quadratic function that can be written in standard form as f(x) = x^2 + 2x - 8.
To graph this function, we can start by finding the vertex, which is located at x = -b/2a = -2/2 = -1, and then plotting a few additional points to determine the shape of the parabola.
To find the y-coordinate of the vertex, we plug x = -1 into the equation and get f(-1) = (-1)^2 + 2(-1) - 8 = -7. So the vertex is at (-1, -7).
We can also find the x-intercepts by setting f(x) = 0 and solving for x:
(x-2)(x+4) = 0
x-2 = 0 or x+4 = 0
x = 2 or x = -4
So the x-intercepts are at (2, 0) and (-4, 0).
To find the y-intercept, we set x = 0 and get f(0) = (0-2)(0+4) = -8. So the y-intercept is at (0, -8).
Now we can plot these points and sketch the parabola:
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The domain of the function is all real numbers, since there are no restrictions on the values that x can take.
To find the range of the function, we can look at the vertex and see that the lowest point on the parabola is at y = -7. Since the parabola opens upward, the range extends from -7 to infinity:
Range: [-7, ∞)
This Took me a long long time