If segment LN is congruent to segment NP and ∠1 ≅ ∠2, prove that ∠NLO ≅ ∠NPM:
2 answers:
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Answer:
see below
Step-by-step explanation:
The function is written in vertex form. It has a negative vertical scale factor, so you know ...
- the vertex is (2, -4)
- the vertical scale factor is 1/2
- the parabola opens downward
Vertex form is ...
f(x) = a(x -h)^2 +k . . . . . . . . . . vertical scale factor "a", vertex (h, k)
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Since you know the vertex and scale factor, you can plot some points on the graph. I find it convenient to think in terms of units either side of the vertex. These are values of x that would make (x-2)^2 be 1^2, 2^2, 3^2, 4^2 and so on. The vertical scale factor of -1/2 tells you that the y-differences for these points will be -1/2, -4/2, -9/2, -16/2 from the vertex. Then we have ...
vertex: (2, -4)
1 unit either side of the vertex: (1, -4.5), (3, -4.5)
2 units either side of the vertex: (0, -6), (4, -6)
3 units either side of the vertex: (-1, -8.5), (5, -8.5)
4 units either side of the vertex: (-2, -12), (6, -12)
You can plot these points and draw a smooth curve through them. Or, you can let a graphing calculator do it. The result will be similar to that shown below.
