Answer:
9514 1404 393
see attached
Step-by-step explanation:
Reflection over the line x = -1 alters the x-coordinates, but leaves the y-coordinates alone. The image point is as far horizontally from the reflection line as the pre-image point is. Each new x-coordinate is the old one subtracted from twice the x-value of the line of reflection: (x, y) ⇒ (-2-x, y) U(-8, -6) ⇒ U'(6, -6) V(-3, -6) ⇒ V'(1, -6) W(-4, -1) ⇒ W'(2, -1)
Answer:
f(-3) = -20
Step-by-step explanation:
We observe that the given x-values are 3 units apart, and that the x-value we're concerned with is also 3 units from the first of those given. So, a simple way to work this is to consider the sequence for x = 6, 3, 0, -3. The corresponding sequence of f(x) values is ...
34, 10, -8, ?
The first differences of these numbers are ...
10 -34 = -24
-8 -10 = -18
And the second difference is ...
-18 -(-24) = 6
For a quadratic function, second differences are constant. This means the next first-difference will be ...
? -(-8) = -18 +6
? = -12 -8 = -20
The value of the function at x=-3 is -20.
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The attachment shows using a graphing calculator to do a quadratic regression on the given points. The graph can then be used to find the point of interest. There are algebraic ways to do this, too, but they are somewhat more complicated than the 5 addition/subtraction operations we needed to find the solution. (Had the required x-value been different, we might have chosen a different approach.)
