The boundary of the lawn in front of a building is represented by the parabola
y = (x^2) /16 + x - 2
And you have three questions which require to find the focus, the vertex and the directrix of the parabola.
Note that it is a regular parabola (its symmetry axis is paralell to the y-axis).
1) Focus:
It is a point on the symmetry axis => x = the x-component of the vertex) at a distance equal to the distance between the directrix and the vertex).
In a regular parabola, the y - coordinate of the focus is p units from the y-coordinate of the focus, and p is equal to 1/(4a), where a is the coefficient that appears in this form of the parabola's equation: y = a(x - h)^2 + k (this is called the vertex form)
Then we will rearrange the standard form, (x^2)/16 + x - 2 fo find the vertex form y = a(x-h)^2 + k
What we need is to complete a square. You can follow these steps.
1) Extract common factor 1/16 => (1/16) [ (x^2) + 16x - 32]
2) Add (and subtract) the square of the half value of the coefficent ot the term on x =>
16/2 = 8 => add and subtract 8^2 => (1/16) [ (x^2) + 16 x + 8^2 - 32 - 8^2]
3) The three first terms inside the square brackets are a perfect square trinomial: =>
(1/16) [ (x+8)^2 - 32 - 64] = (1/16) [ (x+8)^2 - 96] =>
(1/16) [(x+8)^2 ] - 96/16 =>
(1/16) (x +8)^2 - 6
Which is now in the form a(x - h)^2 + k, where:
a = 1/16 , h = - 8, and k = -6
(h,k) is the vertex: h is the x-coordinate of the vertex, and k is the y-coordinate of the vertex.
=> a = 1/16 => p =1/4a = 16/4 = 4
y-componente of the focus = -6 + 4 = -2
x-component of the focus = h = - 8
=> focus = (-8, -2)
2) Vertex
We found it above, vertex = (h,k) = (-8,-6)
3) Directrix
It is the line y = p units below the vertex = > y = -6 - 4 = -10
y = -10
Answer:
This is not a good model because the residual plot becomes thicker at the end
How much was he paid per hour
Answer:
I will just choosé the second one
Because it is possible to compare sizes In one triangle rather than comparing with another one which we don't reàlly know of its sizes.
