Question

How do I find the vertex of -4.9t^2+32t+2

Im having a hard time factoring out the 4.9

Answer 1

use the formula

x = -b/(2a)

where in this case

b = 32 and a = -4.9

Answer 2

this will give you the x coordinate of the vertex

Answer 3

or in this case, the t coordinate

once you figure that out, you plug it into the expression to find the y value of the vertex

Answer 4

ok make sense. How would I do it using difference of squares

Answer 5

this isn't a difference of squares

i think you mean "completing the square" right?

Answer 6

Yes

Answer 7

well there are 2 basic ways

Answer 8

Im stuck on the decimal

Answer 9

first is using the formula above to find the x coordinate of the vertex, then using that to find the y coordinate of the vertex

that will give you h and k respectively

which you can use to plug into

y = a(x-h)^2 + k

Answer 10

or

you can do it like this

Answer 11

I am trying to get it into vertex form like posted above

Answer 12

-4.9t^2+32t+2

-4.9(t^2+32/(-4.9)t+2/(-4.9))

-4.9(t^2-6.5306122t-0.408163265)

-4.9(t^2-6.5306122t + (6.5306122/2)^2 - (6.5306122/2)^2 -0.408163265)

-4.9( (t^2-6.5306122t + (6.5306122/2)^2 - (6.5306122/2)^2) -0.408163265)

-4.9( (t-(6.5306122/2))^2 - (6.5306122/2)^2 -0.408163265)

-4.9( (t-(6.5306122/2))^2 -10.25406066)

-4.9(t-(6.5306122/2))^2 -4.9*(-10.25406066)

-4.9(t-(6.5306122/2))^2 + 50.244897234

-4.9(t-3.2653061)^2 + 50.244897234

Answer 13

that's the formal way to complete the square...which you can see is very ugly

Answer 14

Don't use the difference method so you don't get in trouble with decimals use the first method.

Answer 15

so I recommend you use the first method

Answer 16

and I made a typo somewhere, not sure where

but you should end up with -4.9*(t-3.265306122)^2+54.24489795

Answer 17

ok

Answer 18

here's how I would do it

Answer 19

x = -b/(2a)

x = -32/(2*(-4.9))

x = 3.265306122

so the x coordinate of the vertex is 3.265306122

Answer 20

plug this into the function to find the y coordinate of the vertex

y = -4.9t^2+32t+2

y = -4.9(3.265306122)^2+32(3.265306122)+2

y = 54.24489795


so the vertex is the point (3.265306122, 54.24489795)

Answer 21

awsome.Thanks

Answer 22

which means (h,k) = (3.265306122, 54.24489795)

---> h = 3.265306122, k = 54.24489795

Answer 23

you're welcome

Answer 24

\begin{align}
-4.9t^2+32t+2 &= (-4.9t^2 +32t) +2 \\
&= -4.9\left(t^2 -\frac{320}{49}t -\frac{25600}{2401} \right) +2 \\
&= -4.9\left(t-\frac{160}{49}\right)^2 +2 +\frac{2560}{49} \\
&= -4.9\left(t-\frac{160}{49}\right)^2 +\frac{2658}{49}
\end{align}

Vertex form : \(y=a(x-h)^2 +k\)
your form: \(\large y= -4.9\left(x-\frac{160}{49}\right)^2 +\frac{2658}{49}\)

Answer 25

The quadratic\[ax^2+bx+c=0\]
has a vertex with x-coordinate at \[x=\frac{-b}{2a}\]
______

The t-coordinate of the vertex of \[-4.9t^2+32t+2=0\] is at \[t=\frac{-32}{2\times-4.9}\approx3.2\]