Which transformations were used to create congruent triangle E’F’G from triangle EFG?
1 answer:
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Answer:
B. -2, 2
Step-by-step explanation:
Apparently, we're to presume that f(x) is the line that is graphed. It has a y-intercept of +1 and a slope (rise/run) of 1/2. Its equation is ...
f(x) = 1/2x +1
If we want points of intersection, we want to solve the equation f(x) = g(x) for the values of x that make it so.
1/2x +1 = √(x +2)
Squaring both sides, we get ...
1/4x² +x + 1 = x +2
1/4x² = 1 . . . . . . . . . . . . subtract x+1 from both sides
x² -4 = 0 . . . . . . . . . . multiply by 4, subtract 4
(x -2)(x +2) = 0 . . . . factor the difference of squares
x = -2, 2 . . . . . . . . . values of x that make the factors zero
The solutions to f(x) = g(x) are x = -2 and x = 2.
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<em>Additional comment</em>
The attached graph shows the x- and y-values at the points of intersection. The solutions to f(x) = g(x) are only the x-values, -2 and 2.
The square of (ax +b) is ...
(ax +b)² = a²x² +2abx +b²
The point-slope equation of a line is ...
y = mx + b . . . . . line with slope m and y-intercept b
