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Find the vertex of each parabola by completing the square. x2?6x+8=yx2-6x+8=y

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Find the vertex of each parabola by completing the square. x2?6x+8=yx2-6x+8=y

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The x- intercepts of a parabola are (-5, 0) and (3, 0). What is the function of the parabola? f(x) = -2(x2 + 2x − 5) f(x) = -2(x2 + 2x − 15) f(x) = -2(x2 − 2x − 15) f(x) = -2(x2 − 2x − 5)

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Just need to substitute (with some patience) x=-5 and see whether you get f(x)=0, and x=3.

Now, a parabola is also f(x) proportional to (x+5)*(x-3) (if you know the solutions, which you do in this particular example). So: x^2 + 2*x – 15. This is the third one. The -2 is there to confuse, any number could be there because in this example since f(x)=0 for x=-5 and x=3, it won’t change 0*number=0.

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A parabola is defined by the equation (x − 5)2 = 12(y + 2). In which direction will the parabola open

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A parabola is defined by the equation (x − 5)2 = 12(y + 2). In which direction will the parabola open

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Find the x-intercepts of the parabola with vertex (-2,8) and y-intercept (0,-4). Write your answer in this form: (x1,y1),(x2,y2). If necessary, round to the nearest hundredth.

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Find the x-intercepts of the parabola with vertex (-2,8) and y-intercept (0,-4). Write your answer in this form: (x1,y1),(x2,y2). If necessary, round to the nearest hundredth.

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Find the vertex of the parabola given by the following equation. y=x^2+10x+33

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Answer:

Vertex of the parabola is ( -5 , 8 ).

 

Step-by-step explanation:

Given:

Equation of parabola,

y = x² + 10x + 33

To find: Vertex of the parabola

Vertex form of the parabola is ,

y = a( x – h )² + k

where, ( h , k ) is the vertex of the parabola.

Consider,

y = x² + 10x + 33

y = x² + 10x + 5² – 5² + 33

y = ( x + 5 )² – 25 + 33

y = ( x + 5 )² + 8

by comparing, we get

Vertex is ( -5 , 8 )

Therefore, Vertex of the parabola is ( -5 , 8 )

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The function h(x) = x2 + 14x + 41 represents a parabola. Part A: Rewrite the function in vertex form by completing the square. Show your work. (6 points) Part B: Determine the vertex and indicate whether it is a maximum or a minimum on the graph. How do you know? (2 points) Part C: Determine the axis of symmetry for h(x). (2 points)

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Answered by answersmine AT 22/10/2019 – 03:50 AM

Given the function h(x)=x^2+14x+41, to solve by completing square we procced as follows;
x^2+14x+41=0
x^2+14x=-41
but;
c=(b/2)^2
and b=14
hence;
c=(14/2)^2=49
substituting the value of c in the expression we get:
x^2+14x+49=-41+49
x^2+14x+49=8
(x+7)^2=8
this can be written in vertex form;
h(x)=a(x-h)^2+k
where:
(h,k) is the vertex;
thus
(x+7)^2=8
h(x)=(x+7)^2-8
hence the vertex will be at the point:
(-7,-8)

Post your answer

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Find the vertex, focus, directrix, and focal width of the parabola. -1/40 x^2=y

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Answer:

the vertex is at (0,0)  

the focus is at (0, -10)  

the directrix is y = 10  

the focal width = |-40| = 40

Step-by-step explanation:

The equation of the parabola :

y =frac{-1}{40}x^2

Which shows that the given parabola is a vertical parabola and the standard form of the vertical parabola is written as :

y =frac{1}{4p}(x-h)^2+k .........(1)

the vertex is at (h,k)  

the focus is at (h,k+p)  

the directrix is y = k−p  

the focal width = |4p|

So, changing the given equation of parabola into standard form we get :

y =frac{1}{4times -10}(x-0)^2+0

On comparing it with equation (1) we get

p = -10 , h = 0 , k = 0

Therefore, the vertex is at (0,0)  

the focus is at (0, -10)  

the directrix is y = 10  

the focal width = |-40| = 40

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If the telescope is 1 m deep and 8 m wide, how far is the focus from the vertex? 1 2 4 16 Find the vertex, focus, directrix, and focal width of the parabola. negative 1 divided by 12 times x squared = y

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First, let me introduce the general equation of the parabola:

(x-h)^2 = +/- 4a(y-k) or (y-k)^2=+/- 4a(x-h), where

(h,k) are the coordinates of the vertex
a is the distance of the vertex to the focus
4a = length of lactus rectum or the focal width

If the equation contains (x-h)^2, then the parabola passes the x-axis twice. Similarly, (y-k)^2 passes the y-axis twice. If the sign is (-), it opens to the left(if y-axis) or downward (if x-axis). If the sign is (+), it opens to the right(if y-axis) or upward (if x-axis). 

The equation of the parabola is -1/12 x^2 = y. Rearranging to the general form:

x^2 = -12y
Therefore, 

-4a = -12
4a = 12
a = 3, and the parabola is facing downwards.

The vertex is (0,0) at the origin.
The focus is (0,-3). Since it is negative, the focus is situated downwards, hence -3.
The directrix is the mirror image of the focus. Hence, it is a line passing +3 on the y-axis. y=3
Focal width is 4a which is equal to 12 units.

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Use the graph below for this question: graph of parabola going through negative 5, 1 and negative 4, negative 4. What is the average rate of change from x = −5 to x = −4?

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Answer:

False

Step-by-step explanation:

Suposse that we are given a function f(x) and a constant value h.

1. Case:

If we take the function g(x)=f(x)+h, then the graph of the function g(x) will be the graph of the funcion f(x) moved up or down.

2.Case:

If we take the function g(x)=hf(x), then the graph of the function g(x) will be the graph of the function f(x) just taller or shorter.

3.Case:

If we take the function g(x)=f(x-h), then the graph of the function g(x) will be the graph of the fuction f(x) moved horizontally.

4. Case:

If we take the function g(x)=f(hx), then the graph of the function g(x) will be tha graph of the function f(x) wither or thiner.

For example:

If we take f(x)=sin(x) and h=2. Then, if we take g(x)=sin(2x) then f(0)=g(0)=0, which means that the graph of the functiction is not moved up or down. However, f(π/2)=sin(π/2)=1 and g(π/2)=sin(π)=0 which gives us a hint that the graph of the function became thiner.

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Find the standard form of the equation of the parabola with a focus at (-2, 0) and a directrix at x = 2.

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Standard form is, hold a sec

x=2 is directix
that means it opens left or right
so we must use
(y-k)²=4p(x-h)
where vertex is (h,k) and p is distance from focus to vertex
also shortest distance from vertex to directix
the shortest distance from focus to directix is 2p
if p>0 then the parabola opens right
if p<0 then pareabola opens left

so
(-2,0) and x=2
the distance is 4
4/2=2
p=2
wait, positive or negative
focus is to the left of the directix so p is negative
p=-2

vertex is 2 to the right of the focus and 2 to the left of directix
vertex is (0,0)

so
(y-0)²=4(-2)(x-0) or
y²=-8x is da equation
not sure what form is standard tho

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The daily profit p in dollars of a company making tables is described by the function upper p left parenthesis x right parenthesis equals negative 5 x squared plus 240 x minus 2475 p(x)=?5 x 2+240x?2475?, where x is the number of tables that are manufactured in 1 day. the maximum profit of the company occurs at the vertex of the parabola. how many tables should be made per day in order to obtain the maximum profit for the? company

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The following equation of parabola is given:

p(x)= – 5 x^2 + 240 x – 2475

where p(x) = y

This is a standard form of the parabola. We need to
convert this into vertex form of equation. The equation must be in the form:

y – k = a (x – h)^2

Where h and k are the vertex of the parabola. Therefore,

y = – 5 x^2 + 240 x – 2475

y = -5 (x^2 – 48 x + 495)

Completing the square:

y = -5 (x^2 – 48 x + 495 + _) – (-5)* _

Where the value in the blank _ is = -b/2

Since b = -48        therefore,

y = -5 (x^2 – 48 x + 495 + 81) + 405

y – 405 = -5 (x^2 – 48 x + 576)

y – 405 = -5 (x – 24)^2

Therefore the vertex is at points (24, 405).

The company should make 24 tables per day to attain maximum
profit.

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Find the gradients of the lines that pass through the point (1,7) and are tangent to the parabola y = (2−x)(1+3x).

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Answer:  The correct option is (C). 10 = square root of the quantity of x minus 8 all squared plus y minus 9 all squared

Step-by-step explanation:  Given that the segment AB has point A located at (8, 9). The distance from A to B is 10 units.

We are to select the correct option that could be used to calculate the coordinates for point B.

Let, (x, y) be the co-ordinates of point B.

According to distance formula, the distance between two points (a, b) and (c, d) is given by

d=sqrt{(c-a)^2+(d-b)^2}.

Therefore, the distance between the points A(8, 9) and B(x, y) is given by

d=sqrt{(x-8)^2+(y-9)^2}.

Since, distance between A and B is 10 units, so

d = 10.

Therefore,

10=sqrt{(x-8)^2+(y-9)^2}.

Thus, the correct statement is

10 = square root of the quantity of x minus 8 all squared plus y minus 9 all squared.

Option (C) is correct.

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Find the standard form of the equation of the parabola with a vertex at the origin and a focus at (0, -4).

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Find the standard form of the equation of the parabola with a vertex at the origin and a focus at (0, -4).

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Derive the equation of the parabola with a focus at (2, –1) and a directrix of y = – .5

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Check the picture below…the parabola, with that focus point and that directrix.. .looks more or less like so

now, notice that, the directrix is above the focus, meaning the parabola is vertical and it opens downwards

the distance “p” is the same distance from the vertex to the directrix, since the vertex is half-way between those two fellows

now the  focus point is at 2,-1 and the directrix up above at +0.5, from -1 to 0.5 over the y-axis, there are 1.5 units, the vertex is half-way through, so 1.5/2 or 0.75

so the vertex is from -1 up 0.75 units, or from 0.5 down 0.75 units, that places it at -0.25, the axis of symmetry is x = 2, so, the vertex is at 2, -0.25

because the parabola is opening downwards, the value for “p” is negative, so, we know the distance “p” is 0.75, but because is an opening downwards parabola, then  p = -0.75

now, let’s plug all those guys in, let’s use for -0.25 then -1/4, and for -0.75 the -3/4, which is just the fraction version

bf textit{parabola vertex form with focus point distance}\\&#10;begin{array}{llll}&#10;(y-{{ k}})^2=4{{ p}}(x-{{ h}}) \\&#10;boxed{(x-{{ h}})^2=4{{ p}}(y-{{ k}}) }\&#10;end{array}&#10;qquad &#10;begin{array}{llll}&#10;vertex ({{ h}},{{ k}})\\&#10;{{ p}}=textit{distance from vertex to }\&#10;qquad textit{ focus or directrix}&#10;end{array}\\

bf -------------------------------\\&#10;(x-{{ h}})^2=4{{ p}}(y-{{ k}}) qquad &#10;begin{cases}&#10;h=2\&#10;k=-frac{1}{4}\&#10;p=-frac{3}{4}&#10;end{cases}&#10;\\\&#10;(x-2)^2=4left( -frac{3}{4} right)left( y-left( -frac{1}{4} right) right)implies (x-2)^2=-3left( y+frac{1}{4} right)&#10;\\\&#10;cfrac{(x-2)^2}{-3}=y+cfrac{1}{4}implies cfrac{(x-2)^2}{-3}-cfrac{1}{4}=y&#10;\\\&#10;-cfrac{1}{3}(x-2)^2-cfrac{1}{4}=y

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Which of the equations below could be the equation of this parabola vertex 0 0. A=1/2 x 2 B.y=-1/2 x 2 X=1/2y2 Dx1/2y2

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So hmm check the picture below, that’s about the circle and the endpoints, but notice, the endpoints make up a segment, namely the diameter of the circle, well…. let’s see how long that is, because, the radius is half the diameter

bf textit{distance between 2 points}\ quad \&#10;begin{array}{lllll}&#10;&x_1&y_1&x_2&y_2\&#10;%  (a,b)&#10;&({{ 7}}quad ,&{{ 3}})quad &#10;%  (c,d)&#10;&({{ 7}}quad ,&{{ -5}})&#10;end{array}qquad &#10;%  distance value&#10;d = sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}&#10;\\\&#10;d=sqrt{(7-7)^2+(-5-3)^2}implies d=sqrt{0+(-8)^2}implies d=8&#10;\\\&#10;textit{the radius is half that, so is }boxed{r=4}

now.. hmmm notice, the midpoint of the diameter, is the center of the circle, let’s check that one out

bf textit{middle point of 2 points }\ quad \&#10;begin{array}{lllll}&#10;&x_1&y_1&x_2&y_2\&#10;%  (a,b)&#10;&({{ 7}}quad ,&{{ 3}})quad &#10;%  (c,d)&#10;&({{ 7}}quad ,&{{ -5}})&#10;end{array}qquad &#10;left(cfrac{{{ x_2}} + {{ x_1}}}{2}quad ,quad cfrac{{{ y_2}} + {{ y_1}}}{2} right)&#10;\\\&#10;left(cfrac{{{ 7+7}}}{2}quad ,quad cfrac{{{ -5}} + {{ 3}}}{2} right)implies left( cfrac{14}{2} , cfrac{-2}{2} right)implies boxed{(7,-1)}

now.. that we know what the center is, and what the radius is, well

bf textit{equation of a circle}\\ &#10;(x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2&#10;qquad &#10;begin{array}{lllll}&#10;center (&{{ h}},&{{ kquad }})qquad &#10;radius=&{{ r}}\&#10;&7&-1&4&#10;end{array}

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How would you determine which x-values to use when creating a table of values to graph a parabola?

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A sample is a small subset of a population, on which startistical analysis is carried out to obtain the characteristics of the population.

Sampling is the process of selecting units (e.g., people, organizations) from a
population of interest to obtain the characteristics of the population.

When drawing a sample, it is important that all the components of the population of interest is adequately represented.

Thefollowing samples are categorised based on whether they fairly represent the population of interest or not.

1.) Measuring the heights of every fiftieth person on the school roster to determine the average heights of the boys in the school.

Here, the population of interest is the boys in the school. Drawing a sample of every fiftieth person on the school roster will contain both boys and girls whereas girls are not needed for the purpose of the survey.

Threfore, the sample does not represent fairly the population of interest.

2.) Calling
every third person on the soccer team’s roster to determine how many of
the team members have completed their fundraising assignment.

Here, the population of interest is the team members, so drawing a sample of every third person on the soccer team’s roster represents fairly the population of interest.

3.) Observing every person walking down Main Street at 5 p.m. one evening to determine the percentage of people who wear glasses.

Here the population of interest is people who wear glasses, though observing people walking down the road might be a good way to drawing this sample, but the sample will be biased because by 5 pm, the sum will be down and the people who wear glasses because of the sun might not have their glasses on again.

So this sample does not fairly represent the population of interest.

4.) Sending
a confidential e-mail survey to every one-hundredth parent in the
school district to determine the overall satisfaction of the residents
of the town.

Here, the population of interest is the residents of the town and not all residents of the town might be a parent.

So, the sample of one-hundredth parent in the school district does not fairly represent the population of interest.

5.) Taking
a poll in the lunch room (where all students currently have to eat
lunch) to determine the number of students who want to be able to leave
campus during lunch.

Here, the population of interest is the students and taking a poll in the lunch room (where all students currently have to eat lunch) fairly represent the population of interest.

Therefore, the samples that fairly represent the population are:
Calling
every third person on the soccer team’s roster to determine how many of
the team members have completed their fundraising assignment.
and
Taking
a poll in the lunch room (where all students currently have to eat
lunch) to determine the number of students who want to be able to leave
campus during lunch.

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Given a vertex (h, k) and a point (x, y) on the parabola, derive the quadratic function: (h, k) = (3, 2) and (x, y) = (10, 1).

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Given a vertex (h, k) and a point (x, y) on the parabola, derive the quadratic function: (h, k) = (3, 2) and (x, y) = (10, 1).

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