4. for each function, write the rule in vertex form and find the coordinates of the max/min point on its graph.
a) f(x)=x^2+12x+11
b) g(x)=x^2-4x+7
c) h(x)=x^2-18x+74
d) j(s)=s^2+2s+5
use the results of task 4 to sove each of these equations.
a) x^2+12x+11=0
b) x^2-4x+7=19
c) x^2-18x+74=13
d) s^2+2s+5=53

'Vertex form', having just googled it, is another name for completing the square, which I have done extensively at school. This involved converting:
f(x) = ax^2 + bx + c --> f(x) = a(x-b)^2 + c

In the completed-square form, (b, c) are the co-ordinates of the vertex (which is the maximum/minimum point). So for part a:

- Here is the original
  f(x) = x^2 + 12x + 11

-  Halve the x-coefficient (the number before x) and use it as -b in the vertex form described above. You must then calculate the square of this number and minus it at the end because when you multiply it out, this is the surplus you will make along with x^2 + 12x:
  f(x) = (x + 6)^2  - 36 + 11

- Tidy this up by collecting the constant (just number) terms together:
  f(x) = (x + 6)^2 - 25

- Using the form a(x - b)^2 + c, we can work out that b = -6 (because -b = 6) and c = -25. This gives us the vertex (-6, -25) which is a minimum point because the graph is a positive quadratic, giving us the characteristing 'U' shape which has a bottom. If it were a negative quadratic (denoted by a negative x^2 coefficient), the vertex will be a maximum point because it has an 'n' shape instead.

- To solve the equation from here, make the function equal to zero:
  (x + 6)^2 - 25 = 0

- Then take 25 to the other side:
  (x + 6)^2 = 25

- Next, square-root both sides:
  x + 6 = ±5

- Rearrange to finish:
  x = -6 ± 5 = -1 or -11

- Therefore, the roots (solutions) to the equation x^2 + 12x + 11 = 0 are x = -1 or -11

This method will work the same for the other equations up there too, so I will leave them for you to do.

I hope this helps

In all honesty, there really is no “work” per say to show. mainly it’s just using the process of elimination. the question is asking which domain(s) would be the most appropriate for the function c(n). n is the number of observed vehicles so right away, 1, 2, and 3 can be eliminated.why? 1, 2 start in the negatives and therefore it is impossible to observe negative cars. so, #4 would be the best answer to the given problem.