If you write a quadratic #f(x)=ax^2+bx+c#, you can find the #x# coordinate of the vertex using the formula #(-b)/(2a)#. Let's plug our numbers into this to find the #x#.
#"-7"/"2(-9)" rArr "-7"/"-18" = 7/18#
Next, we plug #x# into #f(x)# to find the #y# of our vertex.
#f(7/18)=-9(7/18)^2+7(7/18)+9#
#f(7/18)=13 13/36#
So your vertex is #(7/18,13 13/36)#
For your zeroes, we can use the quadratic formula.
#x_(1,2) = (-b+-sqrt(color(blue)(b^2-4ac)))/(2a)#
Let's first solve our discriminate, the numbers marked in blue above.
#color(blue)(b^2-4ac) rArr (7)^2-4(-9)(9) = 373#
Now we can plug this into our quadratic formula, along with the other numbers.
#x_(1,2) = (-(7)+-sqrt(373))/(2(-9))#
#x_1 = (-7 + sqrt(373))/-18 ~~ - .684#
#x_2 = (-7 - sqrt(373))/-18 ~~ 1.462#
