1. The zeros of the quadratic function f(x) = (x+6)^2 - 9 can be found by setting f(x) equal to zero and solving for x.
(x+6)^2 - 9 = 0
Taking the square root of both sides:
x+6 = ±√9
x+6 = ±3
Solving for x:
x = -6 ± 3
So the zeros of the quadratic function f(x) are x = -9 and x = -3.
Therefore, the correct answer is A. The zeros are (-3,0) and (-9,0).
2. The vertex form of a quadratic function can be written as f(x) = a(x-h)^2 + k, where (h,k) represents the vertex of the parabola.
In the given function f(x) = 2(x-2)^2 - 2, the vertex is located at (2, -2).
Since the coefficient "a" is positive (2), the parabola opens upwards and the vertex represents the minimum value of the function.
Therefore, the correct answer is A. The function has a minimum value of -2.
3. The y-intercept of a function is the point where the graph of the function crosses the y-axis. In the given quadratic function f(x) = a(x-h)^2 + k, the y-intercept occurs when x = 0.
Plugging in x = 0, we have:
f(0) = a(0-h)^2 + k
f(0) = ah^2 + k
Therefore, the y-intercept of f(x) is (0, ah^2 + k).
So the correct answer is A. (0, ah^2+k).
1. Determine the zeros of the quadratic function f(x)=(x+6)^2−9
A. The zeros are (−3,0) and (−9,0)
B. The zeros are (0,−3) and (0,−9)
C. The zeros are (3,0) and (9,0)
D. The zeros are (3,0) and (−15,0)
2. 2(x−2)^2−2 is the vertex form of the quadratic function f(x). Does this function have a minimum or maximum value and what is it?
A. The function has a minimum value of -2.
B. The function has a minimum value of 2.
C. The function has a maximum value of -2.
D. The function has a maximum value of 2.
3. Let f(x)=a(x−h)^2+k. Where is the y-intercept of f(x)?
A. (0, ah^2+k)
B. (−ah^2+k, 0)
C. (ah^2+k, 0)
D. (0, −ah2+k)
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