Question

How do you find the vertex and intercepts for `x^2-10x-8y+33=0`?

Answer 1

`y_("intercept")=33/8`

Vertex`->(x,y)=(5,1)`

NO X-INTERCEPT

Explanation:

Moving `8y` to the other side of the = and change its sign

`x^2-10x+33=+8y`

To get `y` on its own divide both sides by 8

`(x^2)/8-10/8 x+33/8=y`

Write as:

`y=(x^2)/8-10/8 x+33/8`

Write as:

`y=1/8(x^2color(red)(-10)x)+33/8`

`color(green)(y_("intercept")=+33/8 larr" read directly off the equation")`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(green)(x_("vertex")=(-1/2)xx(color(red)(-10)) =+5)`
The above line is part of the process of completing the square.

`color(green)("The "x^2/8"is positive so the graph is of general shape "uu)`

Substituting `x=+5` gives:

`color(green)(y_("vertex")=1/8[color(white)(./.)5^2-10(5)color(white)(.)]+33/8 =1)`

`color(green)("Vertex"->(x,y)=(5,1))`

As the graph is of general shape `uu` and `y_("vertex")` is above the x-axis there is NO X-INTERCEPT