To answer this question, we need two quadratic functions whose vertices have a vertical distance of 18 units from each other. A quadratic function in vertex form is written as:
f(x) = a(x - h)^2 + k
where `(h, k)` is the vertex of the parabola, `a` is a non-zero coefficient which determines the parabola's direction and width.
Let's assume the vertex of the function `w` is `(h_w, k_w)`. The vertex of function `q` will then be `(h_q, k_q)` such that `k_q = k_w - 18` (since `q` is 18 units below `w`), and we can assume `h_q = h_w` for simplicity, though it's not strictly necessary for the vertical distance condition to hold.
Function `w`:
w(x) = a_w*(x - h_w)^2 + k_w
Function `q`:
q(x) = a_q*(x - h_q)^2 + k_q
= a_q*(x - h_w)^2 + (k_w - 18)
Now we'll need to choose specific values for the constants `a_w`, `h_w`, `k_w`, `a_q`, and `h_q`. To satisfy the condition that the vertices have a vertical distance of 18 units, we just need `k_w` and `k_q` to differ by 18.
For example, if we set:
- `a_w` = 1 (default, just to keep the parabola opening upwards)
- `a_q` = 1 (same as above)
- `h_w` = 0 (any arbitrary point for the vertex of `w`)
- `k_w` = 20 (a value for the vertex of `w`)
- `k_q` = `k_w` - 18 = 2 (vertex of `q` is 18 units below)
Then, the two functions could be:
w(x) = (x - 0)^2 + 20
w(x) = x^2 + 20
q(x) = (x - 0)^2 + 2
q(x) = x^2 + 2
So here, `w` and `q` could be the functions used to create the graphs that are 18 units apart vertically at their respective vertices. Remember that the coefficients `a_w` and `a_q` could be different, and the horizontal placement `h_w` and `h_q` could vary too. As long as the vertical difference in the `k` terms is maintained at 18 units, many sets of functions could satisfy your condition.
Quadratic functions q and w are graphed on the same coordinate grid.
The vertex of the graph of q is 18 units below the vertex of the graph of w.
Which pair of functions could have been used to create the graphs of q and w?
3 answers
Quadratic functions q and w are graphed on the same coordinate grid.
The vertex of the graph of q is 18 units below the vertex of the graph of w.
Which pair of functions could have been used to create the graphs of q and w?
Responses:
q(x) = 18x^2 and w(x) = x^2
q(x) = x^2 + 18 and w(x) = x^2
q(x) = -18x^2 and w(x) = x^2
q(x) = x^2 - 18 and w(x) = x^2
The vertex of the graph of q is 18 units below the vertex of the graph of w.
Which pair of functions could have been used to create the graphs of q and w?
Responses:
q(x) = 18x^2 and w(x) = x^2
q(x) = x^2 + 18 and w(x) = x^2
q(x) = -18x^2 and w(x) = x^2
q(x) = x^2 - 18 and w(x) = x^2
In quadratic functions, the term that affects the vertical position of the vertex is the constant term, when written in standard form:
f(x) = ax^2 + bx + c
The vertex form of a quadratic equation is:
f(x) = a(x - h)^2 + k,
where (h, k) is the vertex of the parabola.
If we want the vertex of q to be 18 units below the vertex of w, we need to make sure that the k value for q is 18 less than the k value for w when both equations are in vertex form.
Let's consider the given response options:
1. q(x) = 18x^2 and w(x) = x^2
- The vertices of both functions are at the origin (0,0) since there is no horizontal or vertical shift. The vertex of q is not 18 units below w.
2. q(x) = x^2 + 18 and w(x) = x^2
- The vertex of q is (0, 18), and the vertex of w is (0, 0). Here, the vertex of q is actually 18 units above w, not below.
3. q(x) = -18x^2 and w(x) = x^2
- Again, the vertices of both functions are at the origin (0,0). The vertex of q is not 18 units below w.
4. q(x) = x^2 - 18 and w(x) = x^2
- The vertex of q is (0, -18), and the vertex of w is (0, 0). In this case, the vertex of q is exactly 18 units below the vertex of w, which matches the given condition.
So, the correct pair of functions that could have been used to create the graphs of q and w, with the vertex of q being 18 units below the vertex of w, is:
q(x) = x^2 - 18
w(x) = x^2
f(x) = ax^2 + bx + c
The vertex form of a quadratic equation is:
f(x) = a(x - h)^2 + k,
where (h, k) is the vertex of the parabola.
If we want the vertex of q to be 18 units below the vertex of w, we need to make sure that the k value for q is 18 less than the k value for w when both equations are in vertex form.
Let's consider the given response options:
1. q(x) = 18x^2 and w(x) = x^2
- The vertices of both functions are at the origin (0,0) since there is no horizontal or vertical shift. The vertex of q is not 18 units below w.
2. q(x) = x^2 + 18 and w(x) = x^2
- The vertex of q is (0, 18), and the vertex of w is (0, 0). Here, the vertex of q is actually 18 units above w, not below.
3. q(x) = -18x^2 and w(x) = x^2
- Again, the vertices of both functions are at the origin (0,0). The vertex of q is not 18 units below w.
4. q(x) = x^2 - 18 and w(x) = x^2
- The vertex of q is (0, -18), and the vertex of w is (0, 0). In this case, the vertex of q is exactly 18 units below the vertex of w, which matches the given condition.
So, the correct pair of functions that could have been used to create the graphs of q and w, with the vertex of q being 18 units below the vertex of w, is:
q(x) = x^2 - 18
w(x) = x^2
1 answer