Question

Given the equation f(x)=2x^2-14x+10, find g(x), the image of f(x) after a ry=x (reflection over the line y=x). Express your answer as a single fraction. ​

Answer 1

Given:

The function is:

`f(x)=2x^2-14x+10`

The function g(x), the image of f(x) after a
`r_(y=x)` (reflection over the line y=x).

To find:

The function g(x).

Solution:

We have,

`f(x)=2x^2-14x+10`

Substitute
`f(x)=y` in the given function.

`y=2x^2-14x+10`

The function g(x), the image of f(x) after a
`r_(y=x)` (reflection over the line y=x). So, interchange x and y.

`x=2y^2-14y+10`

Now, we need to find the value of y.

`x=2(y^2-7y+5)`

`(x)/(2)=y^2-7y+5` [Divide both sides by 2]

`(x)/(2)-5=y^2-7y` [Subtract 5 from both sides]

Add both sides half of square of coefficient of y, i.e.
`((-7)/(2))^2`, to make it perfect square.

`(x)/(2)-5+((-7)/(2))^2=y^2-7y+((-7)/(2))^2`

`(x)/(2)-5+(49)/(4)=y^2-7y+((7)/(2))^2`

`(x)/(2)-5+(49)/(4)=\left(y-(7)/(2)\right)^2`
`[\because (a-b)^2=a^2-2ab+b^2]`

`(x)/(2)+(49-20)/(4)=\left(y-(7)/(2)\right)^2`

Taking square root on both sides, we get

`\pm\sqrt{(x)/(2)+(29)/(4)}=y-(7)/(2)`

`(7)/(2)\pm\sqrt{(2x+29)/(4)}=y`

`(7)/(2)\pm(√(2x+29))/(2)=y`

`(7\pm √(2x+29))/(2)=y`

Substituting
`y=g(x)`, we get

`(7\pm √(2x+29))/(2)=g(x)`

Therefore, the required function is
`g(x)=(7\pm √(2x+29))/(2)`.