1.express in standard form Y+ -3(x+1)^2+5?

Question

2. write in vertex form Y= X^2-12X+40

3. solve by completing the square X^2+6X-7=0

4.solve the equation 2X^2+8X-14=0


please write the number of the question before the answer
thank you and please help

Answer 1

1. Standard form is y = ax^2 + bx + c, so you have to expand the term in parentheses, and then combine like terms:

y = -3(x + 1)^2 + 5
y = -3(x^2 + 2x + 1) + 5
y = -3x^2 - 6x - 3 + 5
y = -3x^2 - 6x + 2

2. Vertex form is the form you were given in (1). In this case, you have to complete the square to get it done:

y = x^2 - 12x + 40
y = (x^2 - 12x + 36) + 40 - 36
y = (x - 6)^2 + 4

3. Completing the square is the method used above for (2):

x^2 + 6x - 7 = 0
x^2 + 6x = 7
x^2 + 6x + 9 = 7 + 9
(x + 3)^2 = 16
x + 3 = 4
x = 1

4. You can do this several ways, but using the quadratic formula will give you an exact answer:

2x^2 - 8x - 14 = 0
x^2 - 4x - 7 = 0

x = [-b +/- sqrt(b^2 - 4ac)] / 2a
x = [4 +/- sqrt((-4)^2 - 4(1)(-7))] / 2(1)
x = [4 +/- sqrt(16 + 28)] / 2
x = [4 +/- sqrt(54)] / 2
x = [4 +/- 3sqrt(6)] / 2

Answer 2

1. standard form is Y = 3(x+1)^2 + 5
Y = 3 (x^2 + 2x + 1) + 5
Y = 3x^2 + 6x + 3 + 5
Y = 3x^2 + 6x + 8
2. vertex form (the sign within the brackets must be negative)
Y = 3 (x-(-1))^2 + 5
a = 3 h = -1 k = 5, so the vertex is (h,k) and the sign of the a value tells us whether it is a positive or a negative (maximum or minumum vaue)
Vertex = (-1,5) and this is a minimum point since a is positive

3. X^2 + 6X - 7 = (x-7)(x+1)
--> how to do this... remember FOIL (First, Outside, Inside, Last). Completing the squares is just breaking it down into its components, so you must start by placing two sets of brackets...
( ) ( )
Then you first look at the squared value. In this case there is no number before the X... so it is a 1 and to get 1... you mulitple 1 by 1, so just place an x at the start of each bracket
(x ) (x )
If say it was 16x^2... then it would be 4x in each bracket... since 4 times 4 is 16... and so on

Next we take a look at the last number (7), and determine all the combinations that can be multiplied together to get 7. In this example 7 and 1 are the only 2, so that makes it easier on us. So... since we know its 7 and 1... we place the 7 and the 1 in the brackets like so...
(x 7) (x 1)
Finally we look at the sign before the 7 and in this case it is a negative. This shows us that the signs must equal a negative (+/+ = +, +/- = -, -/+ = -, -/- = +), so we know that one must be a negative and one must be a positive. To find out which one... look at the middle number... is it a positive or a negative. If it is a positive... the larger number gets the positive sign..
(x + 7) (x - 1) and to check it... just expand it using FOIL
multiply the first values in each bracket together... x*x = x^2, then the outside two values...x * -1 = -1x, then the inside two values.... 7 * x = 7x, then finally the last two... 7 * -1 = -7. Put them together and you get:
X^2 - 1x + 7x - 7 ==> x^2 + 6x - 7

4. For this one you do the same as above.... but i forgot to mention... if you cant factor it out... you must use the quadratic equation. Im sure if you are taking the class you will have that formulas handy, so I won't type it out here. Just plug the numbers into the formula then solve.
ax^2 + bx + c = 0
a = 2, b = 8, c = -14

(-8 +/- 13 4/15) / 4
x = (-21 4/15) / 4 or (5 4/15) / 4

hope that helps

Answer 3

that's 3x -y +2 =0. you basically subtract the y from the left over to the staggering, making it a adverse. leave the two as a great regardless of the shown fact that, considering the fact which you do no longer would desire to do something with it. Cheers - desire this helped =)