Question

write an equation in vertex form of a parabola with zeros of -2 and 5 passing through (0,-15)

Answer 1

please help!!!

Answer 2

alright please note that the zeros are the x intercepts right ?

Answer 3

yes

Answer 4

so x-intercept form is just the factored form of standard form

Answer 5

ok...

Answer 6

alright so the zeros right do you remember how people get the zeros for an equation

Answer 7

nope if i knew how to do the problem i would've done it myself

Answer 8

alright I'm gonna have to teach you everything I know about solving polynomials with example problems get ready to get a math overload. REPEAT MATH OVERLOAD!! All people within a forty mail radius will be busted with math soon. You will thank me for this later. I not just gonna give the solution I goona teach it to you MUWHA!MUWHAHAHAHAHA!!

Answer 9

hahaha ok...:) this sounds interesting..

Answer 10

DO you know how to factor ?

Answer 11

i used to its been two years since I've had to

Answer 12

Now have to give you a problem that you have to try to factor atleast try okay .

Answer 13

haha ok...im not the brightest lightbulb but ok

Answer 14

Nah everyone is smart just had bad teachs all you need is a little moviation to do the problem and don't give ma negative attiude go solve this problem thinking your the smartest person the world.

Answer 15

haha alright

Answer 16

\[x^{2} + 10x+ 25\]

Answer 17

Now factor and tell me how you did it

Answer 18

its (x+5)(x+5)

Answer 19

Do you know what the zeros of that problem is ?

Answer 20

i did (x+ ?) (x+?)
and then i just plugged in the numbers that fit... those were fairly easy

Answer 21

-2 and 5

Answer 22

(x+ 2)(x-5) - 15

Answer 23

There's a helpful hint now put it in standard form.

Answer 24

so thats in standard form?

Answer 25

whys the 15 on the end there?

Answer 26

Cause this is an equation it say it passing through - 15 that's the y intercept its always at the end the the problem the y intercept in when the x is at o and its just y

Answer 27

but I need you to multiply (x+2)(x-5) for me to get standard form

Answer 28

wait but i need vertex form..

Answer 29

\[x^{2} - 3x-15\]

Answer 30

add-15 to the equation

Answer 31

so x^2 - 3x - 30

Answer 32

now use this formula to find the x

Answer 33

pinknabastak, do you have Microsoft Excel on your computer?

Answer 34

yes

Answer 35

and yes once you find the x plug it in the the equation and get the why and I'll explain from there wait can we do this on microsoft excel tell me how

Answer 36

You can easily check your solution on Microsoft Excel. I'll show you how, but first, I'll show you how to find the answer

Answer 37

Recall that a parabola is of the general form \[ax^2 + bx + c = y\]
We are given three points: two zeros and a point of intersection: (-2, 0), (5, 0), and (0, -15)

Answer 38

its not nessicarly finding an answer but simply writing an equation

Answer 39

yes thats all true

Answer 40

That means we can plug those three points into the general formula to generate three separate equations to generate the equation of your parabola (from which you can then get the vertex form). Doing so, we get these three equations:

Answer 41

For (-2, 0):
\[4a - 2b + c = 0\]
For (5, 0):
\[24a + 5b + c = 0\]
For (0, -15):
\[c = -15\]

Answer 42

You then have two equations and two unknowns--you can solve for a and b, then you will get your quadratic equation, from which you can then complete the square to arrive at the vertex form.

Answer 43

You can also plug those three points into Microsoft Excel, plot the points (choose which data set is x, which data set is y). Then generate a polynomial trendline, display the equation, and you will get the plot here: http://i.imgur.com/vwNAFdK.png

Answer 44

And using what rancd had mentioned before, you can find the intercept via the equation
\[x = \frac{-b}{2a}\]
and the corresponding y-value by plugging that x value into your final quadratic equation.

Answer 45

so..... sorry i got kinda lost

Answer 46

You are essentially finding the equation of your parabola by making use of the general quadratic equation
\[f(x) = ax^2 + bx + c = y\]
and the three given points

Answer 47

huh??? so ax^2 +bx -15 = y

Answer 48

Remember, you're given a coordinate, which contains both an x- and a y-value. Applying this idea to the coordinate (0, -15), then you have that x = 0 and y = -15. If you plug this into the general quadratic equation, you get
\[a(0)^2 + b(0) + c = -15\]
\[c = -15\]
Therefore, \[f(x) = ax^2 + bx -15 = y\]

Answer 49

so is that the answer?

Answer 50

No, that is the first step in determining the values of your coefficients a, b, and c. Once you have those, you can complete the square and get the equation in vertex form.

Answer 51

how do i complete the square?

Answer 52

some one help! i just need the equation...

Answer 53

You have two roots: -2 and 5.
That means (x-(-2)) and (x-5) are factors.
So y = a(x+2)(x-5) where a is a constant that we need to find.
(0,-15) is a point on the parabola.
-15 = a(0+2)(0-5)
-15 = -10a
a = -15 / -10 = 3/2

So y = 3/2(x+2)(x-5)
y = 3/2(x^2 -3x - 10)
Put this in vertex form:
y = 3/2 * { (x - 3/2)^2 - (3/2)^2 - 10 } = 3/2 * { (x-3/2)^2 -9/4 - 10 }
y = 3/2 * { (x-3/2)^2 - 49/4 }
y = 3/2(x-3)^2 - 147/8 is the equation of the parabola in vertex form.

Answer 54

wow. Thank you zoo much @ranga its the only thing thats made sense to me in the past hour

Answer 55

Just let me double check for any mistakes in the calculations. Give me a sec.

Answer 56

sorry one more... if you can.... write in factored form of parabola with vertex (5, 18/5) passing through (1, -14/5)

Answer 57

ok

Answer 58

Typo in my last line. It should be: y = 3/2(x-3/2)^2 - 147/8

Answer 59

haha thanks.. what about my other question?

Answer 60

The vertex form of a parabola is: y = a(x-h)^2 + k where (h,k) is the vertex.
They have given you the vertex: (5, 18/5). So h = 5 and k = 18/5
y = a(x-5)^2 + 18/5
It passes through (1, -14/5). Put x = 1, y = -14/5 and find "a".
-14/5 = a(1-5)^2 + 18/5 = 16a + 18/5
16a = -18/5 - 14/5 = -32/5
a= -32/(5*16) = -2/5
So y = -2/5(x-5)^2 + 18/5 = -2/5 * { (x-5)^2 - 9 }
y = -2/5 * { x^2 -10x + 25 - 9 } = -2/5 * { x^2 - 10x + 16 }
y = -2/5(x-8)(x-2)
is the equation of the parabola in factored form.

Answer 61

THANK YOU SOO MUCH!

Answer 62

You are welcome.