3. Identify the vertex and the axis of symmetry of the graph of the function y = 3(x + 2)2 – 3. (1 point)
A- vertex: (2, –3); axis of symmetry: x = 2
B- vertex: (–2, –3); axis of symmetry: x = –2
C- vertex: (2, 3); axis of symmetry: x = 2
D- vertex: (–2, 3); axis of symmetry: x = –2
4. Identify the maximum or minimum value and the domain and range of the graph of the function y = 2(x – 3)2 – 4. (1 point)
A- minimum value: –4
domain: all real numbers
range: all real numbers –4
B- maximum value: 4
domain: all real numbers
range: all real numbers <= 4
C- maximum value: –4
domain: all real numbers <= –4
range: all real numbers
D- minimum value: 4
domain: all real numbers => 4
range: all real numbers
6. Suppose a parabola has vertex (5, –3) and also passes through the point (6, 1). Write the equation of the parabola in vertex form.
(1 point)
y = (x – 5)2 – 3
y = 4(x – 5)2 – 3
y = 4(x + 5)2 – 3
y = 4(x – 5)2 + 3
I don't see your choices, so I can't tell if you did them correctly.
3. The correct answer is A- vertex: (2, –3); axis of symmetry: x = 2. The vertex form of the equation is y = a(x - h)^2 + k, where (h, k) represents the vertex. In this equation, (h, k) = (-2, -3), so the vertex is (2, -3). The axis of symmetry is the vertical line x = h, so the axis of symmetry is x = 2.
4. The correct answer is A- minimum value: -4, domain: all real numbers, range: all real numbers >= -4. The coefficient "a" in the equation y = a(x - h)^2 + k determines whether the parabola opens up or down. Since the coefficient "a" is positive (2), the parabola opens upward, and the vertex represents the minimum point. In this case, the minimum value is -4, the domain is all real numbers, and the range is all real numbers greater than or equal to -4.
6. The correct answer is y = (x - 5)^2 - 3. The equation of a parabola in vertex form is y = a(x - h)^2 + k, where (h, k) represents the vertex. Given that the vertex is (5, -3), we substitute these values into the equation to get: y = (x - 5)^2 - 3.
3. To identify the vertex and the axis of symmetry of the graph of the function y = 3(x + 2)2 – 3, we need to understand the standard form of a quadratic equation, which is given by y = a(x – h)2 + k, where (h, k) represents the vertex of the parabola.
Comparing this with the given equation, we see that a = 3, h = -2, and k = -3. Therefore, the vertex of the parabola is (-2, -3).
The axis of symmetry is a vertical line that passes through the vertex. So, the equation of the axis of symmetry is x = -2.
Therefore, the correct answer is A- vertex: (2, –3); axis of symmetry: x = 2.
4. To identify the maximum or minimum value and the domain and range of the function y = 2(x – 3)2 – 4, we can again use the standard form of a quadratic equation, which is y = a(x – h )2 + k.
Comparing this with the given equation, we see that a = 2, h = 3, and k = -4.
The parabola opens upwards because a positive value of a is multiplied with the squared term. This means that the vertex represents the minimum value of the function.
Therefore, the minimum value of the function is -4.
The domain of the function is all real numbers, as there are no restrictions on the input values (x).
The range of the function is all real numbers greater than or equal to -4, because the parabola is shifted downward by 4 units.
Therefore, the correct answer is A- minimum value: -4; domain: all real numbers; range: all real numbers greater than or equal to -4.
6. To write the equation of the parabola in vertex form, given that the parabola has a vertex at (5, -3) and passes through the point (6, 1), we can use the vertex form of the quadratic equation, which is y = a(x – h)2 + k.
Substituting the vertex coordinates (h, k) = (5, -3), we have y = a(x – 5)2 – 3.
Substituting the point (6, 1) into the equation, we can solve for 'a'.
1 = a(6 – 5)2 – 3
1 = a(1)2 – 3
1 = a – 3
a = 4
Substituting the value of 'a' back into the equation, we get y = 4(x – 5)2 – 3.
Therefore, the correct answer is D- y = 4(x – 5)2 + 3.