Area of equilateral triangle inscribed in a circle with radius 2
1 answer:
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Answer: It's a tie between f(x) and h(x). Both have the same max of y = 3
The highest point shown on the graph of f(x) is at (x,y) = (pi,3). The y value here is y = 3.
For h(x), the max occurs when cosine is at its largest: when cos(x) = 1.
So,
h(x) = 2*cos(x)+1
turns into
h(x) = 2*1+1
h(x) = 2+1
h(x) = 3
showing that h(x) maxes out at y = 3 as well
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Note: g(x) has all of its y values smaller than 0, so there's no way it can have a max y value larger than y = 3. See the attached image to see what this graph would look like if you plotted the 7 points. A parabola seems to form. Note how point D = (-3, -2) is the highest point for g(x). So the max for g(x) is y = -2
The highest point shown on the graph of f(x) is at (x,y) = (pi,3). The y value here is y = 3.
For h(x), the max occurs when cosine is at its largest: when cos(x) = 1.
So,
h(x) = 2*cos(x)+1
turns into
h(x) = 2*1+1
h(x) = 2+1
h(x) = 3
showing that h(x) maxes out at y = 3 as well
--------------------------------
Note: g(x) has all of its y values smaller than 0, so there's no way it can have a max y value larger than y = 3. See the attached image to see what this graph would look like if you plotted the 7 points. A parabola seems to form. Note how point D = (-3, -2) is the highest point for g(x). So the max for g(x) is y = -2