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)
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. To determine the zeros of the quadratic function f(x)=(x+6)^2−9, we set f(x) equal to zero and solve for x:
(x+6)^2−9 = 0
Expanding the square, we have:
x^2 + 12x + 36 - 9 = 0
Simplifying, we get:
x^2 + 12x + 27 = 0
Now we can factor the quadratic equation:
(x + 3)(x + 9) = 0
Setting each factor equal to zero, we have:
x + 3 = 0 or x + 9 = 0
Solving for x, we get:
x = -3 or x = -9
Therefore, the zeros of the quadratic function f(x)=(x+6)^2−9 are (-3, 0) and (-9, 0). Therefore, the correct option is A. The zeros are (-3, 0) and (-9, 0).
2. The given function, 2(x−2)^2−2, is in vertex form, which is in the form f(x) = a(x−h)^2 + k. In this form, the vertex is represented by the coordinates (h, k) and determines whether the function has a minimum or maximum value.
We can see that the equation is in the form f(x) = a(x−h)^2−2, where a = 2, h = 2, and k = -2.
The vertex form implies that the vertex is at (h, k) = (2, -2).
Since the coefficient "a" is positive (2 > 0), the parabola opens upwards and the vertex represents the minimum point of the function.
Therefore, this function has a minimum value of -2. Therefore, the correct option is A. The function has a minimum value of -2.
3. In the vertex form of a quadratic function f(x) = a(x−h)^2 + k, the y-intercept occurs when x = 0. Let's substitute x = 0 into the equation:
f(0) = a(0−h)^2 + k = ah^2 + k
The y-intercept is the point (0, f(0)), so it can be represented as (0, ah^2 + k).
Therefore, the correct option is A. The y-intercept of f(x) = a(x−h)^2 + k is (0, ah^2 + k).
1. To determine the zeros of the quadratic function f(x) = (x+6)^2 - 9, we need to find the values of x for which f(x) equals zero. In other words, we need to solve the equation (x+6)^2 - 9 = 0.
To do this, we can start by adding 9 to both sides of the equation:
(x+6)^2 - 9 + 9 = 0 + 9
(x+6)^2 = 9
Next, we take the square root of both sides of the equation:
√((x+6)^2) = √9
(x+6) = ±3
Now, we can solve for x by subtracting 6 from both sides of the equation:
(x+6) - 6 = ±3 - 6
(x) = ±3 - 6
x = -3 - 6 or x = 3 - 6
Therefore, the zeros of the quadratic function f(x) = (x+6)^2 - 9 are x = -9 and x = 3.
The correct answer is option A: The zeros are (-3, 0) and (-9, 0).
2. The given equation 2(x-2)^2 - 2 is in vertex form, which is of the form f(x) = a(x-h)^2 + k. In vertex form, the vertex of the quadratic function is represented by the point (h, k).
Comparing the given equation with the vertex form equation, we can see that h = 2 and k = -2.
Since the coefficient of the quadratic term (a) is positive (2), the parabola opens upwards, and therefore, the vertex represents the minimum value of the function.
Therefore, the function has a minimum value. To determine the value, we substitute the x-coordinate of the vertex (h = 2) into the equation:
f(2) = 2(2-2)^2 - 2
f(2) = 2(0)^2 - 2
f(2) = -2
So, the function has a minimum value of -2.
The correct answer is option A: The function has a minimum value of -2.
3. The y-intercept of a quadratic function is the point where the graph intersects the y-axis. The y-coordinate of this point will be the value of the function when x equals 0.
For the given quadratic function f(x) = a(x-h)^2 + k, to find the y-intercept, we need to substitute x = 0 into the equation:
f(0) = a(0-h)^2 + k
f(0) = a(-h)^2 + k
f(0) = ah^2 + k
Therefore, the y-intercept of the function f(x) = a(x-h)^2 + k is the point (0, ah^2 + k).
The correct answer is option A: (0, ah^2+k).