If <em>x</em>² + <em>y</em>² = 1, then <em>y</em> = ±√(1 - <em>x</em>²).
Let <em>f(x)</em> = |<em>x</em>| + |±√(1 - <em>x</em>²)| = |<em>x</em>| + √(1 - <em>x</em>²).
If <em>x</em> < 0, we have |<em>x</em>| = -<em>x</em> ; otherwise, if <em>x</em> ≥ 0, then |<em>x</em>| = <em>x</em>.
• Case 1: suppose <em>x</em> < 0. Then
<em>f(x)</em> = -<em>x</em> + √(1 - <em>x</em>²)
<em>f'(x)</em> = -1 - <em>x</em>/√(1 - <em>x</em>²) = 0 → <em>x</em> = -1/√2 → <em>y</em> = ±1/√2
• Case 2: suppose <em>x</em> ≥ 0. Then
<em>f(x)</em> = <em>x</em> + √(1 - <em>x</em>²)
<em>f'(x)</em> = 1 - <em>x</em>/√(1 - <em>x</em>²) = 0 → <em>x</em> = 1/√2 → <em>y</em> = ±1/√2
In either case, |<em>x</em>| = |<em>y</em>| = 1/√2, so the maximum value of their sum is 2/√2 = √2.
Answer:Given that the graph shows tha the functión at x = 0 is below the y-axis, the constant term of the function has to be negative. This leaves us two possibilities:
y = 8x^2 + 2x - 5 and y = 2x^2 + 8x - 5
To try to discard one of them, let us use the vertex, which is at x = -2.
With y = 8x^2 + 2x - 5, you get y = 8(-2)^2 + 2(-2) - 5 = 32 - 4 - 5 = 23 , which is not the y-coordinate of the vertex of the curve of the graph.
Test the other equation, y = 2x^2 + 8x - 5 = 2(-2)^2 + 8(-2) - 5 = 8 - 16 - 5 = -13, which is exactly the y-coordinate of the function graphed.
Step-by-step explanation:
Answer:
Step-by-step explanation:
the formula for a parallelegram is just base * height soo...
n*b*h
