Question

How do you graph `(y-3)^2=8(x-3)`?

Answer 1

`x_("intercept")=35/8=4.375`

The is no `y_("intercept")`

Vertex`->(x,y)=(3,3)`

Explanation:

Assumption: `y` is the dependant variable (answer)

`color(blue)("Deriving an equation that is simpler to plot")`

Multiply out the RHS bracket

`(y-3)^2=8x-24`

Square root both sides

`y-3=sqrt(8x-24)`
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Consider this point demonstrated by example

`(-3)xx(-3)=+9`
`(+3)xx(+3)=+9`

So depending on the conditions the square root of any number, say `a` is `+-sqrt(a)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Using the above principle we have:

`y-3=+-sqrt(8x-24)`

Add 3 to both sides

`y=3+-sqrt(8x-24)->" graph shape of "sub`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the critical points on the plot")`

For the values to remain in the 'real' set of numbers ( `RR` )
`8x-24` must remain positive. The trigger point is when `x<3`.

So for `yinRR` we have `{x: x>=3}`

Thus there is not any plot for `x<3`

`color(brown)("y-intercept")`

Set `x=0` this does not comply with `{x: x>=3}`
so there is no y-intercept.

`color(brown)("x-intercept")`

Set `y=0=3+sqrt(8x-24)`

Thus for this to work `sqrt(8x-14)` needs to be -3. This is not acceptable so this condition does not exist.

Set `y=3-sqrt(8x-24)`

So this means that `8x-24=+9" " =>" " 8x=35`

`x=35/8=4.375`

`color(brown)("Determine "y_("vertex")" at "x_("vertex")=3)`

This is the turning point ( vertex ) of `sub`

`y=3+-sqrt(24-24)=3`

Vertex`->(x,y)=(3,3)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Tony B

`color(blue)("Some people would say that the plot should be:")`

Tony B