I really need your help?

Question

1. Factor

3x^2 - 7xy - 6y^2

2. When a ball is thrown, its height in feet h after t seconds is given by the equation

h = vt - 16t^2

where v is the initial upward velocity in feet per second. If v = 15 feet per second, find all values of t for which h = 3 feet. Round answer to 2 decimal places.

3. What is the x - intercept(s) and the coordinates of the vertex for the parabola y = x^2 +4x-5. If there is more than one x-intercept seperate them with commas.

Thank you

Answer 1

1. unless you have a lot of practice with these you just have to try out the possibilities. the coefficients of x can only be 1 and 3 to give you 3x^2. the coefficients of y could be 1 and 6 or 2 and 3, and one of these coefficients has to be negative to give you -6y^2. you know you have it right when the coefficient of xy is -7.

2. substituting in the values given you have to solve

3 = 15t - 16t^2

or, making one side equal to zero:

16t^2 - 15t - 3 = 0

the fact that you're asked to round the answer indicates you probably have to use the quadratic formula to find the solution since it's not a round number. that is, for the quadratic equation

ax^2 + bx + c = 0

x = (-b +/- sqrt(b^2 - 4ac))/2a = (15 +/- sqrt(33))/32

you can probably work it out from there, remember to round the answer to 2dp

3. you have the x-intercepts when y = 0 so you need to solve

x^2 + 4x - 5 = 0

it's easy to see you can factorise this

(x + 5)(x - 1) = 0

which has solutions of x=-5 and x=1

coordinates of the vertex... hmm ok the vertex i think is the maximum/minimum value of the parabola, which occurs when the derivative is equal to zero:

dy/dx = 2x + 4 = 0

x = -2 (you could also get this by symmetry from the x intercepts i suppose, the vertex is half way between the intercepts)

then you need the y value when x = -2, which comes to -9, so the vertex is at (-2, -9)

Answer 2

1 ) 3x^2 - 7xy - 6y^2

Here, -7 is the coefficient of xy and -18 is the product of the coeficients of x^2 and y^2. Now, think of two numbers whose sum is -7 and product is -18. They are -9 and 2. So break up
-7xy into -9xy and 2xy and proceed as under.

3x^2 - 7xy - 6y^2 = 3x^2 -9xy +2xy -6y^2

= 3x ( x - 3y) + 2y ( x - 3y )

= ( x - 3y ) ( 3x + 2y )

2 ) Given equation is 16t^2 -vt +h = 0

Putting v = 15 and h = 3, 16t^2 -15t + 3 = 0

Solving this quadratic equation gives

t = ( 15 + root 33 )/32 or ( 15 - root 33 ) / 32

= 0.65 sec or 0.29 sec.

3 ) Putting y = 0 in the equation y = x^2 + 4x - 5 and solving for x,

x^2 +4x - 5 = 0 gives x = -5 or x = 1

Hence, x-intercepts are -5, 1

To find the co-ordinates of vertex without calculus, remember that since coefficient of x^2 is positive, vertex is where y has a minimum value.

Now, y = x^2 + 4x -5

= ( x + 2 )^2 - 9 (making perfect square using terms

containing x^2 and x and adjusting
the value of the constant )
Now, y will have a minimum value when x + 2 = 0, or x = -2

and the minimum value of y will be -9

Thus vertex is ( -2, -9 ).

Answer 3

Hey! Here the answers to your questions:
1. Just split it into 2 brackets
=(3x+2y) x (x-3y)

2. First substitute in the values:
3=15t-16t^2
Then make it into one quadratic equation:
16t^2-15t+3=0
Solving gives two answers:
.29, .65

3.To find the x-intercept you have to factorise it into two brackets:
=(x-1) x (x+5)
Thus, 1 and -5 are your intercepts.
To find the coordinates of the vertex you must complete the square:
= x^2+4x+4-9
=(x+2)^2 - 9
thus vertex is (-2,-9)

Answer 4

5,6,2