3 years ago

HURRY Given f(x)=x^2−2x−5. Enter the quadratic function in vertex form in the box.

Mathematics

1 answer:

romanna [79]3 years ago

5 0

Answer:

$f(x)=(x-1)^2-6$

Step-by-step explanation:

The given quadratic function is

$f(x)=x^2-2x-5$

We complete the square to obtain the vertex form;

Add and subtract half the coefficient of $x$

$f(x)=x^2-2x+(-1)^2-(-1)^2-5$

Factor by identifying the perfect square

$f(x)=(x-1)^2-1-5$

Simplify to obtain the vertex form as

$f(x)=(x-1)^2-6$

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On a unit circle, the vertical distance from the x-axis to a point on the perimeter of the circle is twice the horizontal distan

zalisa [80]

Answer:

(2√5)/5

Explanation:

CHECK THE ATTACHMENT BELOW FOR DETAILED EXPLATION

From the question we can deduced that the vertical distance from the x-axis to a point on the perimeter of the circle is twice the horizontal distance, we can say that the y-coordinate, is twice of the horizontal, then let call the horizontal "y", the vertical would be 2y.

from the attachment the question on the unit circle is interpreted using a figure,then, hypotenuse was first calculated