7題MC要教我點做 V

1. Find the max value of the quadratic function y = -x^2 + 3x + 1.
y = -x^2 + 3x + 1
y = -(x^2 - 3x) + 1
y = -[(x)^2 - 2(x)(3/2) + (3/2)^2] + 1 + (3/2)^2
y = -[x - (3/2)]^2 + 13/4

Therefore, the max value is 13/4.

2. Which of the following function has a min value of -1?
The answer is C.
It is because when you put x = -1 into B and D, y is equal to 0 and 1 respectively.
When you put x = 1 into A and C, y is equal to 0 and -1 respectively, therefore, the answer C.

3. Find the axis of symmetry of the graph.
y = a(x + h)^2 + k
When x = -h, y = k, where (-h , k) is the vertex of the graph.
Therefore, the axis of symmetry is x = -h.

4. Find the vertex of the function y = -x^2 + 6x - 3.
y = -x^2 + 6x - 3
y = -(x^2 - 6x) - 3
y = -[(x)^2 - 2(x)(3) + (3)^2] - 3 + 3^2
y = -(x - 3)^2 - 3 + 9
y = -(x - 3)^2 + 6

The vertex of the function is (3 , 6).

5. The new function after reflected in the x-axis is:
The old function : y = x^2 + 4x - 5
When the function, y = f(x), reflected in the x-axis, the f(x) = -f(x)
Therefore, after the transformation, the old function, y = x^2 + 4x - 5 would become y = -(x^2 + 4x - 5), i.e. y = -x^2 - 4x + 5

6. f(x) = x^2 + bx + c
If the function translated 4 units to right, the new function would become:
f(x) = (x^2 + bx + c) + 4 = g(x)
g(x) = x^2 + bx + (c + 4)
g(x) = [(x)^2 + 2(x)(b/2) + (b/2)^2] + (c + 4) - (b/2)^2
g(x) = (x + b/2)^2 + (c - b^2/4 + 4)

As the vertex is (1 , -3), therefore, -b/2 = 1 and c - b^2/4 + 4 = -3
b = -2 and c = -6
therefore f(x) = x^2 - 2x - 6
f(x) = x^2 - 2x - 6
f(x) = [(x)^2 - 2(x)(1) + 1] - 6 - 1
f(x) = (x - 1)^2 - 7
therefore, the axis of symmetry of f(x) is y = -7.

7. If the function, f(x), is translated k unites to left, the function would become f(x + k).
Therefore, as k = +2, which means that the graph is translated to left by 2 units.
As a result, the answer is A.