f(x)=x^2−6

Replace f(x) with yy.

y=x^2−6

Interchange the variables.

x=y2−6

Solve for yy.

Move −6-6 to the right side of the equation by subtracting −6-6 from both sides of the equation.y2=6+xy2=6+xTake the squaresquare root of both sides of the equationequation to eliminate the exponent on the left side.y=±√6+xy=±6+x

The complete solution is the result of both the positive and negative portions of the solution.

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y=√6+x,−√6+x
Solve for y and replace with f^−1(x).

Answer is f−1(x)=√6+x,−√6+x

3 0

3 years ago

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Vertex of f(x) = 1/2(x-2)^2-3

AlekseyPX

Vertex is (2,-3). H (which is 2) sign is always opposite and k stays the same

3 0

3 years ago

Reference attached image for the problem.Please show your work for finding the partial area.

-Dominant- [34]

To find the area of a sector of a circle use the next formula:

$A=\frac{\theta}{360º}*\pi *r^2$

As the given circle has a outside angle 90º (it is not part of the sector of the circle) subtract the 90º from 360º (total angle of a circle) to find the angle of the sector:

$\theta=360º-90º=270º$

Find the area of the sector with angle 270º:

$\begin{gathered} A=\frac{270º}{360º}*3.14*(11in)\placeholder{⬚}^2 \\ \\ A=0.75*3.14*121in^2 \\ \\ A=284.955in^2 \end{gathered}$ Then, the approximate area of the given sector of a circle is 284.955 square inches