Question

Can somebody help me? I need help writing the equations for the parabolas in the graph attached. I have given the vertex.

Answer 1

here is the graph

Answer 2

@Compassionate @amistre64 @ganeshie8 @paki @iGreen @mathmale @thomaster @iambatman @zepdrix @Zarkon

Answer 3

now work the roots

Answer 4

personally i work the roots, then adjust the vertex

Answer 5

okay so how would I do that given the information above?

Answer 6

what are the roots to a given parabola?

Answer 7

a root is also known as an x intercept

Answer 8

it is the values of x when y=0

Answer 9

I have the intersection points of x being -4 and -1 for the first parabola and 1 and 4 for the second

Answer 10

lets work on the one with roots 1,4

now, by definition, f(root) = 0 and this can be accomplished by multiplication

1-x=0 when x=1
4-x = 0 when x=4

therefore

f(x) = a(1-x)(4-x) is a suitable start.

solving for a is simple enough by using the vertex

Answer 11

Okay so would I split them up and put
f(x)=a(1-x)
f(x)=a(4-x)

Answer 12

then solve?

Answer 13

no need to split anything up ... not sure why you would even consider that

f(x) = a(1-x)(4-x) is the function, the equation, its just in a useful form and not a polynomial

Answer 14

okay so that is the equation.

Answer 15

when x=1, f(1)=0

f(1) = a(1-1)(4-1)
= a(0)(3) = 0




when x=4, f(4)=0

f(1) = a(1-4)(4-4)
= a(-3)(0) = 0

Answer 16

the vertex is when x is the middle of 1 and 4, so let x=3.5 and f(3.5) is equal to about 4.3 you say soo

4.3 = a(1-3.5)(4-3.5) to solve for a

Answer 17

a(-1.25)

Answer 18

4.3 = a(-1.25)

now divide off to find a

Answer 19

a=3.44

Answer 20

-3.44

Answer 21

@amistre64

Answer 22

good, so our function for one of them is:\[f(x)=-3.44(1-x)(4-x)\]
if we notice that the other is just a shift to the left by 4 we have the equation of the other one

Answer 23

a shift by 5 that is :)

Answer 24

f(x)=-3.44(1-(x-5))(4-(x-5))

Answer 25

ugh, i shifted that to the right ... not the left

Answer 26

f(x)=-3.44(1-(x+5))(4-(x+5)) thats better

Answer 27

now, your problem does not ask for a polynomial, just an equation ... so this would suffice.

Answer 28

So that fits both parabolas?

Answer 29

one is for one, the other is for the other

Answer 30

okay so the equations are
\[f(x)=-3.4(1-x)(4-x)\]
\[f(x)=-3.44(1-(x+5))(4-(x+5))\]

Answer 31

essentially yes ... -3.44 for the top as well

simplify them to your hearts content

Answer 32

oh yeah sorry I mistyped

Answer 33

I'm sorry but I got something totally different by just using the vertex. I started with the one on the left, and since the vertex is at (-2.5, 4.3), the center is translated to \[(x+2.5)^{2}=-4(y-4.3)\]Solving for p in the focal point you get that p = -1 (because expanding that you get -4y + 17.2 and 4(p)=-4 so p must equal -1) so that would put the focus at (-2.5, 3.3) and the directrix would be the line y = 5.3.

Answer 34

But that's just my interpretation.

Answer 35

when x=1 or 4, that does not equal 0

Answer 36

your process is flawed at some point

Answer 37

in the geometric interpretation, the is a correction factor usually called p such that:\[y=4px\]

Answer 38

lol, y=4px^2

Answer 39

somethings off with that memory, but its still valid

Answer 40

Okay so @amistre64 how can we turn the equations you gave me into quadratic equations?
\[f(x)=ax^2+bx+c\] ?

Answer 41

@amistre64

Answer 42

@IMStuck @iGreen @precal can somebody help me take the equations amistre64 gave me and turn them into quadratic equations?

Answer 43

I did this yesterday and unfortunately did not save my papers but first off, I found that the high point of the graph, which is at y = 4.3 occurs when x = -2.5. I'm not sure where amistre got -3.44. I did it a different way just using the vertex. Let me re-do and then post. Ok?

Answer 44

you cant have a quadratic equation with 2 vertexes

Answer 45

and we already worked it out, just exapnd what we got

Answer 46

-3.44 is the multiplier that fixes the roots to the vertex.

there is a family of parabolas that have the roots (1-x)(4-x) they all differ by their vertex, which is just some multiplier value such that: f(x) = a(1-x)(4-x)

a is solved when we use the vertex for x and f(x)

Answer 47

http://www.wolframalpha.com/input/?i=y%3D-3.44%281-x%29%284-x%29%2Cy%3D-3.44%281-%28x%2B5%29%29%284-%28x%2B5%29%29%2C+x%3D-5..5%2C+y%3D-5..5

not quite the best syntax for wolfram, but notice that the graph is 2 arches.

Answer 48

to demonstrate a family of parabolas that have roots 1,4

http://www.wolframalpha.com/input/?i=y%3D%281-x%29%284-x%29%2Cy%3D3%281-x%29%284-x%29%2Cy%3D-5%281-x%29%284-x%29%2Cy%3D-%281-x%29%284-x%29%2F3%2C

notice that they all differ by some multiplier that determines their vertex.

Answer 49

i do have a miscalculation when solving for a :)

Answer 50

Rewriting it from the formula\[(x-h)^{2}=-4(p)(y-k)\]I just inserted the vertex like this:\[(x+2.5)^{2}=-4(y-4.3)\]

Answer 51

4.3 = -2.25a , i just assumed the -1.25 was correct

a = -1.911

Answer 52

why do you assume p=1?

Answer 53

I can't remember how I worked it...I'm trying to redo it now...getting kinda messed up!

Answer 54

You do your thing and I will try to follow as well...

Answer 55

I actually assumed p = -1 because the parabola has a max, not a min.

Answer 56

you can solve for p by using a root

\[(1+2.5)^{2}=-4p(0-4.3)\]
\[\frac{(3.5)^{2}}{-4(-4.3)}=p\]

Answer 57

you cant assume a value of 1

Answer 58

p comes out to .7122 now

Answer 59

since p=.1308 approx

your setup gives

(x-2.5)^2 = -4(.1308)(y-4.3)

\[y=\frac{(x-2.5)^2}{-4(.1308)}+4.3\]

http://www.wolframalpha.com/input/?i=y%3D%5Cfrac%7B%28x-2.5%29%5E2%7D%7B-4%28.1308%29%7D%2B4.3

Answer 60

sorry, im working the right parabola

Answer 61

OMG!!! Wow! That's the problem!!!!!!!

Answer 62

after i work the right parabola, i just shifted it to the left by 5

Answer 63

in your left, work the roots as -1,0 not 1,0 and youll get the same p as me

Answer 64

When you were solving for p up above, why did you put a + sign in with the (1 + 2.5)^2 and a negative with the (0 - 4.3). Please clarify.

Answer 65

I need two quadratic equations. 1 for each parabola Then I have instructions to solve those for specifics to include in the final product I will turn in. But I do not know how to convert the equations amistre gave me

Answer 66

i had used your left side setup, and used a right side root by mistake

Answer 67

the root 1,0 is for a right side vertex of x-2.5, y-4.3

the root -1,0 is for a left side vertex of x+2.5, y-4.3

i corrected it afterwards in my head

Answer 68

When I do the math with a + inside the (x - h)^2, I still get that p = .7122. I used your set up and did not get p = .1308

Answer 69

I am now also working on the right parabola, not the left.

Answer 70

use the left side root

Answer 71

then your vertex setup is in error

Answer 72

Please show us.

Answer 73

tell me the vertex

Answer 74

AydenCarpenter14, we will help you. Promise.

Answer 75

the vertex is at (2.5, 4.3) for the parabola on the right.

Answer 76

the equation i gave you, which i was off in a, are just addition/multipication to exapnd them

Answer 77

I did this and got your p value...\[(1-2.5)^{2}=-4p(0-4.3)\]

Answer 78

ok, so for that specific vertex, we use the root 1,0 or 4,0; i chose 1,0

the setup for that vertex is:\[(x-2.5)^2=4p(y-4.3)\]as compared to your original setup what used the other vertex intstead becuase you had typed it as:\[(x+2.5)^{2}=4p(y-4.3)\]

Answer 79

Are you back on the left now?

Answer 80

Let's do the right. Saves confusion.

Answer 81

i demonstarted that you presented us with a left side setup to start with

Answer 82

I gotcha. But let's do the right one first. K?

Answer 83

then sticking to the right side vertex setup, and the stated root from the right as 1,0
\[(1-2.5)^2=4p(0-4.3)~:~p=-1.911...\]

Answer 84

Ok, the left side of that comes to 2.25. Yes?

Answer 85

Right side comes to 0 - 17.2p, yes?

Answer 86

4(4.3) = 17.2 yes

Answer 87

I mean 0 + 17.2p

Answer 88

But the 4 is negative! Isn't it!?

Answer 89

the p will adapt to whatever your sign needs to be

Answer 90

4p gives us p = -1.911

-4p gives us p=1.911

so lets just stick to the setup given to avoid confusion

Answer 91

But you still get that when you divide 2.25 by 17.2 it is .1308, not -1.911

Answer 92

How did you get that value of p? Please clarify! This is important stuff!

Answer 93

For the right parabola, I have the set up as\[(1-2.5)^{2}=-4p(0-4.3)\]Is that correct?

Answer 94

(1-2.5)^2=4p(0-4.3)

(-1.5)^2=p(-17.2)

your right, i was working ahead of myself :)

Answer 95

Ok, so the p = .1308. Yes?

Answer 96

i had originally stated: "since p=.1308 approx"

and then expanded it out to the quadratic to get my a = -1.911 and confused the 2 in the process.