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)
aswer please
1. The zeros of the quadratic function f(x)=(x+6)^2-9 are (-3, 0) and (-9, 0). The correct answer is A.
2. The function 2(x-2)^2-2 does have a minimum value, which is -2. The correct answer is A.
3. The y-intercept of the function f(x)=a(x-h)^2+k is (0, ah^2+k). The correct answer is A.
1. To find the zeros of a quadratic function, we need to set the function equal to zero and solve for x.
f(x) = (x+6)^2 - 9
Setting f(x) equal to zero:
0 = (x+6)^2 - 9
Adding 9 to both sides:
9 = (x+6)^2
Taking the square root of both sides:
±√9 = x+6
Simplifying:
±3 = x+6
Subtracting 6 from both sides:
x = -3-6 = -9
Therefore, the zeros of the quadratic function are x = -3 and x = -9.
The correct answer is A. The zeros are (-3, 0) and (-9, 0).
2. The vertex form of a quadratic function is f(x) = a(x-h)^2 + k, where the vertex of the parabola is (h, k).
In the given function, f(x) = 2(x-2)^2 - 2, we can identify the vertex as (2, -2).
Since the coefficient of the squared term (a) is positive, the parabola opens upwards, resulting in a minimum value.
Therefore, the function has a minimum value of -2.
The correct answer is A. The function has a minimum value of -2.
3. In the equation f(x) = a(x-h)^2 + k, the y-intercept is the point where x = 0.
Substituting x = 0 into the equation, we can find the y-coordinate of the y-intercept.
f(0) = a(0-h)^2 + k
Simplifying:
f(0) = a(-h)^2 + k
f(0) = ah^2 + k
Therefore, the y-intercept of f(x) is at the point (0, ah^2 + k).
The correct answer is A. The y-intercept is (0, ah^2 + k).
To determine the zeros of a quadratic function, we need to find the x-values where the function equals zero. In this case, we have the quadratic function f(x) = (x+6)^2 - 9.
1. To find the zeros of the function f(x), we need to set f(x) equal to zero and solve for x:
(x+6)^2 - 9 = 0
Now, let's expand and simplify the equation:
(x+6)^2 - 9 = 0
(x+6)(x+6) - 9 = 0
(x^2 + 12x + 36) - 9 = 0
x^2 + 12x + 27 = 0
Now we have a quadratic equation in general form: ax^2 + bx + c = 0, where a = 1, b = 12, and c = 27.
We can solve this equation by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:
The quadratic formula is given by: x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values a = 1, b = 12, and c = 27 into the quadratic formula, we get:
x = (-12 ± √(12^2 - 4(1)(27))) / (2(1))
x = (-12 ± √(144 - 108)) / 2
x = (-12 ± √36) / 2
x = (-12 ± 6) / 2
So the two possible solutions for x are:
x = (-12 + 6) / 2 = -6 / 2 = -3
x = (-12 - 6) / 2 = -18 / 2 = -9
Therefore, the zeros of the quadratic function f(x) = (x+6)^2 - 9 are x = -3 and x = -9.
The answer is A. The zeros are (-3, 0) and (-9, 0).
2. The vertex form of a quadratic function is given by f(x) = a(x-h)^2 + k, where the vertex of the parabola is (h, k).
In this case, we have the vertex form function f(x) = 2(x-2)^2 - 2.
The coefficient in front of the squared term tells us whether the parabola opens upward (if positive) or opens downward (if negative). Since the coefficient is positive (2), the parabola opens upward.
Since the parabola opens upward, the vertex represents the minimum point of the function. The y-coordinate of the vertex gives us the minimum value.
Comparing the vertex form f(x) = 2(x-2)^2 - 2, we can see that the vertex is (2, -2).
Therefore, the function has a minimum value of -2.
The answer is A. The function has a minimum value of -2.
3. In the general form of the quadratic function f(x) = a(x-h)^2 + k, the y-intercept is the point where the graph crosses the y-axis. To find the y-intercept, we set x = 0 and solve for y.
Plugging x = 0 into the function, we get:
f(0) = a(0-h)^2 + k
f(0) = ah^2 + k
So the y-intercept is given by the point (0, ah^2 + k).
The answer is A. The y-intercept of f(x) is (0, ah^2 + k).