Quadratics help? Thanks!?

Question

1. Use the Completing the Square method to find the vertex form of the quadratic function y = 2x^2 + 8x + 18.


2. In the question, use the information s(t) = -16t^2 + v0t + s0

We throw a rock into the air with initial velocity of 50 ft/sec, and initial position of 6 ft. Find how high the rock goes before coming back down. To answer this question, fill in the blanks in the following sentence, where the first blank gives this height, and the second blank gives how many seconds into the flight of the rock it reaches this maximum height.
The rock reaches its highest position of _______ feet above the ground after _______ seconds of flight

These are some select problems that I'm not sure about out of a bunch that I get from the math from random worksheets I downloaded to teach myself Alg 2 over the summer. I numbered them for the convenience of answering. Quadratics are odd. If anyone can solve all of them it would be great, because I learn best from seeing it done... Thanks again!

Answer 1

Question Number 1 :

For this function y(x)= 2*x^2 + 8*x + 18 , answer the following questions :

A. Find the x-intercept (Function) !

B. Find the y-intercept (Function) !

C. Find the minimum/maximum point of the function !



Answer Number 1 :

The equation 2*x^2 + 8*x + 18 = 0 is already in a*x^2+b*x+c=0 form.

In that form, we can easily derive that the value of a = 2, b = 8, c = 18.



1A. Find the x-intercept (Function) !

The coordinate where the curve y = 2*x^2 + 8*x + 18 intersect with x-axis,

will always be in the form of ( x , y ) = ( x, 0 )

So the next thing we have to do is subtituting y with 0 in y = 2*x^2 + 8*x + 18

Then we have to solve the equation : 2*x^2 + 8*x + 18 = 0

We have to use completing the square to solve the problem

2*x^2 + 8*x + 18 = 0 ,divide both side with 2

So we get x^2 + 4*x + 9 = 0 ,

The coefficient of x is 4

We have to use the fact that ( x + q )^2 = x^2 + 2*q*x + q^2 , and assume that q = 4/2 = 2

Which means we can turn the equation into x^2 + 4*x + 4 + 5 = 0



Since the constant in the right side is positive, the Discriminant value of this equation should be negative.

The function y = 2*x^2 + 8*x + 18 will not intersect x-axis in real Cartesian space.

Therefore the said function dont have x-intercept in real Cartesian space.

Unless you are asked to find the x-intercept in the complex Cartesian space, you can stop here.



Which is the same with ( x + 2 )^2 + 5 = 0

And it is the same with (( x + 2 ) - 2.23606797749979*i ) * (( x + 2 ) + 2.23606797749979*i ) = 0

And it is the same with ( x + 2 - 2.23606797749979*i ) * ( x + 2 + 2.23606797749979*i ) = 0

The function have 2 x-intercept in complex Cartesian coodinate

They are ( x , y ) = ( -2 + 2.23606797749979*i , 0 ) and ( x , y ) = ( -2 - 2.23606797749979*i , 0 )



1B. Find the y-intercept (Function) !

For the curve y = 2*x^2 + 8*x + 18 to intersect with y-axis,

We have to remember that y-axis itself is a line with equation x = 0

So the next thing we have to do is subtituting x with 0 in y = 2*x^2 + 8*x + 18

So we get y = 2*(0)^2 + 8*0 + 18

Which can be turned into y = 18

So the function have y-intercept in ( x , y ) = ( 0 , 18 )



1C. Find the minimum/maximum point of the function !

Since the value of a = 2 is positive, the function y(x) = 2*x^2 + 8*x + 18 have a minimum point.

Use the formula y'(x) = 0 , to find the value of x in the minimum point

We have to find the function y'(x) first

So we get y'(x) = 4*x + 8 = 0

Which means that 4*x = -8

Which means that x = -8/4

So we get x = -2

So the minimum point is ( x , y ) = ( -2 , y(-2) )

Which is ( x , y ) = ( -2 , 10 )





Question Number 2 :

For this function s(t)= -16*t^2 + 50*t + 6 , answer the following questions :

A. Find the t-intercept (Function) !

B. Find the minimum/maximum point of the function !



Answer Number 2 :

The equation -16*t^2 + 50*t + 6 = 0 is already in a*x^2+b*x+c=0 form.

In that form, we can easily derive that the value of a = -16, b = 50, c = 6.



2A. Find the t-intercept (Function) !

The coordinate where the curve s = -16*t^2 + 50*t + 6 intersect with t-axis,

will always be in the form of ( t , s ) = ( t, 0 )

A way to find value of x and y that match both equations is by subtituting s with 0

So we have to solve : -16*t^2 + 50*t + 6 = 0

We have to use completing the square to solve the problem

-16*t^2 + 50*t + 6 = 0 ,divide both side with -16

Then we get t^2 - 3.125*t - 0.375 = 0 ,

The coefficient of t is -3.125

We have to use the fact that ( t + q )^2 = t^2 + 2*q*t + q^2 , and assume that q = -3.125/2 = -1.5625

By using that fact we turn the equation into t^2 - 3.125*t + 2.44140625 - 2.81640625 = 0

And it is the same with ( t - 1.5625 )^2 - 2.81640625 = 0

And it is the same with (( t - 1.5625 ) - 1.67821519776219 ) * (( t - 1.5625 ) + 1.67821519776219 ) = 0

By opening the brackets we will get ( t - 1.5625 - 1.67821519776219 ) * ( t - 1.5625 + 1.67821519776219 ) = 0

So we just need to add up, and get ( t - 3.24071519776219 ) * ( t + 0.115715197762194 ) = 0

So the function have 2 t-intercept in ( t , s ) = ( 3.24071519776219 , 0 ) and ( t , s ) = ( -0.115715197762194 , 0 )



2B. Find the minimum/maximum point of the function !

Since the value of a = -16 is negative, the function s(t) = -16*t^2 + 50*t + 6 have a maximum point.

Use the formula s'(t) = 0 , to find the value of t in the maximum point

We have to find the function s'(t) first

So we get s'(t) = - 32*t + 50 = 0

Which means that -32*t = -50

Which means that t = -50/-32

So we get t = 1.5625

So the maximum point is ( t , s ) = ( 1.5625 , s(1.5625) )

Which is ( t , s ) = ( 1.5625 , 45.0625 )





The highest position 45.0625 feets

( the s-value in in answer 2B )

After 1.5625 seconds

( the t-value in in answer 2B )



It will fall back to the ground after 3.24071519776219 seconds

( the positive t-intersect in in answer 2A )

Answer 2

I can do one...I think....



2x^2+8x= -18 move constant

x^2+4x= -9 get rid of the 2 with the square by dividing

x^2 + 4x + 4 = -9+8 because of the 2 it would be multiplied by

(x+2)^2 = -1 take 1/2 the first term, sign of second, square of third term

x+2=i because can't take square of negative 1 it gets an i with the square root

x= -2+i or -2-i





for the second, I would put it in a calculator. If you don't have one u can download one. v=50 s=6 i think...

the equation would be y=-16x+50x+6 and find the max or highest point on the graph

Answer 3

the answer to question a million is: y=2(x+2)(x+a million) y = 2x^2 + 6x + 4 y=2(x^2 + 3x + 2) y=2(x+2)(x+a million) Are you useful that query 2 isn't y = x^2 - 4 because of the fact the answer to that must be (x-2)(x+2) If that's no longer then the question has no answer!