The image of a parabolic lens is projected onto a graph. The image crosses the x-axis at –2 and 3. The point (–1, 2) is also on
the parabola. Which equation can be used to model the image of the lens?
y = -1/5 (x – 2)(x + 3)
y = -1/3(x – 2)(x + 3)
y = -1/2 (x + 2)(x – 3)
y = 1/4 (x + 2)(x – 3)
y = -1/5 (x – 2)(x + 3)
y = -1/3(x – 2)(x + 3)
y = -1/2 (x + 2)(x – 3)
y = 1/4 (x + 2)(x – 3)
2 answers:
For x-intercepts "b" and "c", a quadratic can be written as
y = a(x -b)(x -c)
You know the x-intercepts, so you can write the equation as
y = a(x +2)(x -3) . . . . . . eliminates the first two answer choices
Substituting (x, y) = (-1, 2), you have
2 = a(-1+2)(-1-3) = -4a
Then dividing by -4 gives
-2/4 = a = -1/2 . . . . . corresponds to the third answer choice
The appropriate selection is
y = -1/2 (x + 2)(x - 3)
y = a(x -b)(x -c)
You know the x-intercepts, so you can write the equation as
y = a(x +2)(x -3) . . . . . . eliminates the first two answer choices
Substituting (x, y) = (-1, 2), you have
2 = a(-1+2)(-1-3) = -4a
Then dividing by -4 gives
-2/4 = a = -1/2 . . . . . corresponds to the third answer choice
The appropriate selection is
y = -1/2 (x + 2)(x - 3)
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