F(x) = x³ - 7
f(x + 2) = (x + 2)³ - 7
f(x + 2) = (x + 2)(x + 2)(x + 2) - 7
f(x + 2) = (x(x + 2) + 2(x + 2))(x + 2) - 7
f(x + 2) = (x(x) + x(2) + 2(x) + 2(2))(x + 2) - 7
f(x + 2) = (x² + 2x + 2x + 4)(x + 2) - 7
f(x + 2) = (x² + 4x + 4)(x + 2) - 7
f(x + 2) = (x²(x + 2) + 4x(x + 2) + 4(x + 2)) - 7
f(x + 2) = (x²(x) + x²(2) + 4x(x) + 4x(2) + 4(x) + 4(2)) - 7
f(x + 2) = (x³ + 2x² + 4x² + 8x + 4x + 8) - 7
f(x + 2) = (x³ + 6x² + 12x + 8) - 7
f(x + 2) = x³ + 6x² + 12x + 8 - 7
f(x + 2) = x³ + 6x² + 12x + 1
g(x) = f(x + 2) + 4
g(x) = (x³ + 6x² + 12x + 1) + 4
g(x) = x³ + 6x² + 12x + 1 + 4
g(x) = x³ + 6x² + 12x + 5
<h3>
Answer: Choice B) purple and orange</h3>
The purple graph reflects over the line y = x to get the orange graph. Every point on the purple graph reflects over y = x using the rule (x,y) --> (y,x). So the x and y values swap places more or less.
We do not reflect over any horizontal line, which is why the blue and purple don't pair up as inverse pairs.
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The blue doesn't pair with the orange because of two reasons
1) The first local max, or highest hill, on the blue graph means that for some small region/interval, the y value is highest. However, the first bump in the orange graph shows that x is smallest for some y interval. The x and y swap when it comes to inverses.
2) The blue graph has y go off to negative infinity for both endpoints. So x must go off to negative infinity for the inverse (again x and y swap), but this isn't the case. Instead x goes off to positive infinity for both endpoints in the orange graph.
Answer:
See attached picture for example, hope it helps!
Step-by-step explanation:
A*C = B, so vecAB = vecBC, vecAB = (-1/2+3, -3) and vecBC(xc+1/2, yc-4)
it means -1/2+3 = xc+1/2, -3=yc-4, so C(2, -1)
