Bear in mind that, when it comes to trigonometric functions, the location of the exponent can be a bit misleading, however recall that sin²(θ) is really [ sin( θ )]²,
![\bf 2sin^2(2x)=2\implies sin^2(2x)=\cfrac{2}{2} \\\\\\ sin^2(2x)=1\implies [sin(2x)]^2=1\implies sin(2x)=\pm\sqrt{1} \\\\\\ sin(2x)=\pm 1\implies sin^{-1}[sin(2x)]=sin^{-1}(\pm 1)](https://tex.z-dn.net/?f=%5Cbf%202sin%5E2%282x%29%3D2%5Cimplies%20sin%5E2%282x%29%3D%5Ccfrac%7B2%7D%7B2%7D%0A%5C%5C%5C%5C%5C%5C%0Asin%5E2%282x%29%3D1%5Cimplies%20%5Bsin%282x%29%5D%5E2%3D1%5Cimplies%20sin%282x%29%3D%5Cpm%5Csqrt%7B1%7D%0A%5C%5C%5C%5C%5C%5C%0Asin%282x%29%3D%5Cpm%201%5Cimplies%20sin%5E%7B-1%7D%5Bsin%282x%29%5D%3Dsin%5E%7B-1%7D%28%5Cpm%201%29)
Write an equation where x
1 answer:
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Answer:
Step-by-step explanation:
Vertex form is accomplished by completing the square on the quadratic. Do this by first setting the parabola equal to 0 then moving the constant over to the other side:
Now take half the linear term, square it, and add it to both sides. Our linear term is 6. Half of 6 is 3, and 3 squared is 9:
The reason we do this is to create a perfect square binomial on the left:
(obviously the 0 results from the addition of 9 and -9). Move the 0 back over to the other side and set the quadratic back equal to y:
This gives you a vertex of (-3, 0), which is a minimum value, since the parabola opens upwards.
