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?
To answer this question, we should take into account what the vertex of a graph of a quadratic function represents and how it is calculated. A quadratic function commonly takes the form:
f(x) = ax^2 + bx + c
The vertex of a quadratic function is the highest or lowest point on the graph, depending on whether the parabola opens upwards (a > 0) or downwards (a < 0). The x-coordinate of the vertex is given by -b/(2a), and the y-coordinate can be found by substituting this back into the function f(x).
Let's denote the given functions by q(x) and w(x) and their vertices as V(q) and V(w), respectively.
Let's say w(x) has a vertex V(w) = (h, k). The function q(x) will have a vertex V(q) = (h, k - 18) because it is said to be "18 units below the vertex of w".
The pair of functions would differ only in their c value (the y-intercept), because this difference in the constant term would affect the vertical position of the graph, moving the vertex up or down without altering the parabola's shape or the x-coordinate of the vertex.
If w(x) is given by:
w(x) = ax^2 + bx + c
Then the function q(x) would be:
q(x) = ax^2 + bx + (c - 18)
The a and b coefficients would remain the same because the shapes of the graphs are the same (same direction of opening and same width), but the constant term c in q(x) would be 18 less than that in w(x) to reflect the shift downward.
As a concrete example, if:
w(x) = 2x^2 - 4x + 6 (with a randomly chosen vertex above the x-axis)
Then q(x) could be:
q(x) = 2x^2 - 4x - 12 (which shifts the vertex of w(x) down by 18 units)