Question 730578
x squared and y to the power of 1; the equation is a parabola.


For standard form, solve for y as a function of x and then complete the square for x.  Do you want more complete help?  


"Yes", so then...


INCLUDING SOME DETAILS:

Conic sections will have one or more variables in degree 2, with variables 
to degree no higher than 2.  This is how you know your equation is a conic
section.  Parabolas have one variable to degree 2 and the other variable (if  
it's a two-variable equation) to degree 1.  This then is how we know your
equation is a parabola.



Now, some of the process to start toward standard form.
{{{x^2 - 14x - 9y + 22 = 0}}}



Solve for y.
{{{x^2-14x+22=9y}}}, by just adding +9y to both sides
{{{y=(1/9)x^2-(14/9)x+22/9}}}
{{{y=(1/9)x^2-(1/9)*14x+22/9}}}
We want an expression in x with coefficient of 1 to which we
may Complete the Square...
{{{y=(1/9)(x^2-14x)+22/9}}}


Complete The Square.
Now we need to use the term, {{{(14/2)^2=49}}} and ADD it and SUBTRACT
it to the right-hand side... You need to study to know why and how that value
was picked...
{{{y=(1/9)(x^2-14x+49)+22/9 -49/9}}}, be sure you understand this step perfectly.
{{{y=(1/9)(x-7)^2-27/9}}}
{{{highlight(y=(1/9)(x-7)^2-3)}}}, DONE!