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:
5/(3x+2) = 7/(5x-2)
<=>5*(5x-2) = 7*(3x+2) (x is not equal to -3/2 or 2/5)
<=> 25x -10 = 21x +14 (x is not equal to -3/2 or 2/5)
<=> 4x = 24 (x is not equal to -3/2 or 2/5)
<=> x = 6 (valid)
Answer:
Step-by-step explanation:
so, this is a quadratic equation, meaning two solutions, and we have a factored form of it, meaning you can get the solutions by simply zeroing out the f(x).
![\bf \stackrel{f(x)}{0}=-(x-3)(x+11)\implies 0=(x-3)(x+11)\implies x= \begin{cases} 3\\ -11 \end{cases} \\\\\\ \boxed{-11}\stackrel{\textit{\large 7 units}}{\rule[0.35em]{10em}{0.25pt}}-4\stackrel{\textit{\large 7 units}}{\rule[0.35em]{10em}{0.25pt}}\boxed{3}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7Bf%28x%29%7D%7B0%7D%3D-%28x-3%29%28x%2B11%29%5Cimplies%200%3D%28x-3%29%28x%2B11%29%5Cimplies%20x%3D%20%5Cbegin%7Bcases%7D%203%5C%5C%20-11%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20%5Cboxed%7B-11%7D%5Cstackrel%7B%5Ctextit%7B%5Clarge%207%20units%7D%7D%7B%5Crule%5B0.35em%5D%7B10em%7D%7B0.25pt%7D%7D-4%5Cstackrel%7B%5Ctextit%7B%5Clarge%207%20units%7D%7D%7B%5Crule%5B0.35em%5D%7B10em%7D%7B0.25pt%7D%7D%5Cboxed%7B3%7D)
so the zeros/solutions are at x = 3 and x = -11, now, bearing in mind the vertex will be half-way between those two, checking the number line, that midpoint will be at x = -4, so the vertex is right there, well, what's f(x) when x = -4?
![\bf f(-4)=-(-4-3)(-4+11)\implies f(-4)=7(7)\implies f(-4)=49 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{vertex}{(-4~~,~~49)}~\hfill](https://tex.z-dn.net/?f=%5Cbf%20f%28-4%29%3D-%28-4-3%29%28-4%2B11%29%5Cimplies%20f%28-4%29%3D7%287%29%5Cimplies%20f%28-4%29%3D49%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20%5Cstackrel%7Bvertex%7D%7B%28-4~~%2C~~49%29%7D~%5Chfill)
Answer:
<R=23°
Step-by-step explanation:
The sum of angles in a triangle is 180°
So, <R + <S + <T = 180°
<R+90°+67°=180°
<R+157°=180°
<R=23°
