Math Geniuses. Pls help me out with these problem.? Pls Explain your steps.?

Question

1) I need the coordinates of five points including the vertex and intercepts. y= 10x - x^2 -16

2) Factor by completing the square x+ 6x^2 -2.

I NEED a good explanation on comlpeting the square number 2

Answer 1

1.

y = 10x - x^2 - 16

First, rearrange it in descending powers of x.

y = -x^2 + 10x - 16

Factor out a -1, effectively switching all the signs but most importantly making -x^2 into x^2

y = (-1)(x^2 -10x + 16)

Now, factor the quadratic. Note that -2 and -8 multiply to make 16, and add up to make -10, so those are the two numbers you choose.

y = (-1)(x - 8)(x - 2)

To get the x-intercepts, you make y equal to 0 (zero).
0 = (-1)(x-8)(x-2)

You can divide by -1 to get rid of it

0 = (x-8)(x-2)
Which means x-8 = 0 and x - 2 = 0
Thus, x = 8, x = 2,
so the x-intercepts are 2 and 8.

To solve for the vertex, refer back to the original question in descending x powers order

y = -x^2 + 10x - 16

Your goal here is to complete the square. First, make x^2 positive, but pull out -1 ONLY for the first two terms.

y = -(x^2 - 10x) - 16

Now, the trick to completing the square is to "add zero" to the function. The key value we need is "half-squared" of the number in front of x. This value is important for completing the square, so let's calculate that.

By "half-squared", I mean, take half of the value, and then square it. In our example, it's -10. Take half of that, and it's -5. Squared, and it's 25. So 25 is our key value.

Now, we're going to ADD 25 within the brackets AND balance it out in order to make sure we added zero.

y = -(x^2 - 10x + 25) - 16 + ?????

Note that we just inserted 25 in the brackets. However, outside the brackets there's a negative sign, so we really inserted -25 in there. How do we balance it out? By adding 25 where the question marks should be.

y = -(x^2 - 10x + 25) - 16 + 25
y = -(x^2 - 10x + 25) + 9

Now, we factor. When we factor, it *SHOULD* be a square function, given the reason why we added 25 in there.

y = -(x - 5)^2 + 9

In the general equation y = A(x-h)^2 + k, the coordinates of the vertex are (h,k). In the above case, we take the negative of -5 as our h (which is 5), and this is our x-coordinate. We take the 9 as our k, and this is the y-coordinate.

Therefore, the coordinates of the vertex is (5,9)

2. To complete the square of

x + 6x^2 - 2

First, rearrange in descending power

6x^2 + x - 2

Pull out the 6 for the first two terms, since, for completing the square, it is absolutely essential that x^2 is left by itself.

6(x^2 + 1/6 x) - 2

Now, we used the "Half squared" Method we used above. In this case, we're taking half-squared of 1/6. Half of 1/6 is 1/12, squared is 1/144. So that's our magic number.

6(x^2 + 1/6 x + 1/144) - 2 + ??????

Now, to fill in our question marks again. Note that there's a 6 outside the brackets, so we essentially really added in 6 times 1/144, which is equal to 6/144. We have to offset this by subtracting 6/144, so that's what we put in the question marks.

6(x^2 + 1/6 x + 144) - 2 - 6/144

Let's change 2 into 288/144

6(x^2 + 1/6 x + 144) - 288/144 - 6/144
6(x^2 + 1/6 x + 144) - 294/144

And then now, we factor the square.
6(x + 1/2)^2 - 294/144

Answer 2

uhhh does x^2 equal x squared, if so i might be able to help you...

first: organize it
- x^2 + 10x -16 = y
second: get rid of the - in front of x^2, by multiplying -1 to all numbers
x^2 -10x + 16 = -y
third: make y = 0
x^2 - 10x + 16 = 0
fourth: solve for x (should get two points that way)
(x-8) = 0 (x-2) = 0
x = (8,0) (2,0)
To find the axis of symetry : a.o.s.= - b/2a
-(-10) / 2(1) = 10/2 = 5 (b = -10, and a =1)
so your A.O.S.= 5, that is also your x in your vertex
so you now plug it in for x^2 - 10x + 16 = 0
you then get (5)^2 - 10(5) +16 = 0
which is 25 - 50 + 16 = - 9 , so now you know your y in your vertex, it's -9
now to find the y intercept, use the original equation
y = 10x - x^2 -16, and substitute all the x's for 0
y = 0 - 0 -16, now you have another point (0,-16)
[so now you have 4 points, (8,0), (2,0), (5,-9), (0, -16) now i kinda forgot how to find the 5 points, but 4 seems ok :) ]

Factoring:
first organize
6x^2 + x - 2 = 0
and then use the quadratic formula
-b +/- Square root of (b^2 -4ac)
over 2 a
which is -(1) +/- square root of (1)^2 - 4(6)(-2) / 2(6) = -1 +/- square root of (1 + 48) / 12
which becomes -1 +/- square root of (49) / 12
which equals
(-1 +/- 9 )/ 12
then get (-1 + 9) / 12 and (-1 - 9) / 12
which you get 8/12 and -10/12
then do put (x + 8/12) (x - 10/12) or you can simplify it to...
(x + 2/3) (x - 5/6)
(i think i'm pretty sure i'm right on both, but there might be a possibility i did something wrong)

Answer 3

1)
y = 10x - x^2 - 16
y = -1( x^2 - 10x + 16)
y = -1(x^2 - 10x + 25 + 16 - 25)
(y - 9) = - (x - 5)^2
The vertex is (5, 9)
y-intercept is -16
x intercepts are 2 and 8
y(1) = 10 - 1 - 16 = -7

2)
x + 6x^2 - 2 =
6(x^2 + (1/6)x - 1/3) =
6(x^2 + (1/6)x + (1/12)^2 - 1/3 - (1/12)^2) =
6(x + 1/12)^2 - 1/3 - 1/144) =
6(x + 1/12)^2 - (48 + 1)/144) =
6(x + 1/12)^2 - (7/12)^2 =
6(x + 1/12 + 7/12)(x + 1/12 - 7/12) =
6(x + 8/12)(x - 6/12) =
6(x + 2/3)(x - 1/2) =
(3x + 2)(2x - 1)

Answer 4

1) pick five numbers and substitute 'x' in the equation for the numbers. i.e x=5

10 x 5 - 5^2 - 16 =
50 - 25 - 16 =
9

so the first coordinate would be 5 (x), 9 (y)

do this five times with five different numbers and you have five cordinates. use close numbers like 5,7,9,11,13.

plot them on a graph.

bang

Answer 5

1) the vertex (5;9) the x of the vertex equals -b/a (-10/-2)
The intercepts (2;0) and (8;0) you equal the funtion to 0, you may use the formula
[-b(+/-)(sqrt)b`2-4*a*c]/2*a

Other points (1;-7) and (3;5)
you choose randomly to values of x and you replace them in the equation

Answer 6

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