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

that's 3x -y +2 =0. you in basic terms subtract the y from the left over to the superb, making it a unfavorable. leave the two as an excellent nevertheless, in view which you do no longer would desire to do something with it. Cheers - wish this helped =)