identify the vertex and the axis of symmetry of the graph for the function y=3(x+2)^2
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
identify the maximum or minimum value and the domain and range of the graph of the function y=2(x-3)^2-4.
a.minimum value -4
domain all real numbers
range all real numbers_>_-4
b.max value 4
domain all real numbers
range all real numbers_<_4
c. max value -4
domain all real numbers_<_-4
range all real numbers
d. minimum value 4
domain all real numbers_>_4
range all real numbers
for y = 3(x+2)^2
the vertex is (-2,0) and the axis of symmetry is x = -2
According to the choices, you probably forgot to add/subtract at the end.
if it is y = 3(x+2)^2 + c
then the vertex is (-2,c)
for the 2nd, y = 2(x-3)^2 - 4
the vertex is (3,-4) and it opens upwards,
So the minimum value is -4
the domain is any real number
and the range is y ≥-4
Here is how Wolfram graphed the 2nd, all you have to do is change the equation to graph the first.
http://www.wolframalpha.com/input/?i=y%3D2%28x-3%29%5E2-4
To identify the vertex and the axis of symmetry of the graph for the function y = 3(x+2)^2, you can follow these steps:
Step 1: Recall that the vertex form of a quadratic function is given by y = a(x-h)^2 + k, where (h, k) represents the vertex of the parabola.
Step 2: Compare the given function y = 3(x+2)^2 with the general form.
In this case, the value of a is 3, h is -2, and k is 0 (since there is no value added at the end).
Step 3: Use the values of h and k to identify the vertex coordinates.
The vertex is given by the formula (h, k). So, in this case, the vertex would be (-2, 0).
Step 4: Identify the axis of symmetry.
The axis of symmetry is a vertical line that passes through the vertex. In this case, the axis of symmetry is x = -2.
Therefore, the correct answer is:
a. vertex (-2, 0) and axis of symmetry x = -2.
For the second question, to identify the maximum or minimum value, as well as the domain and range of the function y = 2(x-3)^2-4, follow these steps:
Step 1: Recognize that the given function is in vertex form, y = a(x-h)^2 + k.
Here, a is 2, h is 3, and k is -4.
Step 2: Use the vertex form to determine the coordinates of the vertex.
The vertex is given by the formula (h, k). Thus, the vertex of this function is (3, -4).
Step 3: Determine whether the parabola opens upward or downward.
Since the coefficient 'a' is positive (2 is positive), the parabola will open upward.
Step 4: Identify the maximum or minimum value.
In this case, the vertex is the minimum point because the parabola opens upward. The minimum value of the function is -4.
Step 5: Determine the domain and range.
The domain is the set of all possible x-values. For quadratic functions, the domain is always all real numbers (-∞, +∞).
The range is the set of all possible y-values. Since the minimum value is -4 and the parabola opens upward, the y-values start from -4 and go to positive infinity (-4, +∞).
Therefore, the correct answer is:
a. minimum value -4, domain all real numbers, range all real numbers ≥ -4.