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Help please, if you know how to do it. Posting separate questions so you get more points,l thanks

Help please if you know how to do it Posting separate questions so you get more pointsl thanks class=
Help please if you know how to do it Posting separate questions so you get more pointsl thanks class=

Respuesta :

9514 1404 393

Answer:

  1. y = x^2 +x +1; (-1/2, 3/4)
  2. y = 3x^2 -15x +12; (-2, 54), (5, 12)
  3. y = -1/2x^2 +2x +5; (0, 5)

Step-by-step explanation:

1. The graph looks like the vertex is at x = -1/2. So we can start with the equation y = (x +1/2)^2 +k. At x=0, this becomes 1 = 1/4 +k, or k = 3/4. The leading coefficient of 1 is consistent with the other points shown, so the equation is ...

  y = (x +1/2)^2 +3/4 . . . . vertex is (-1/2, 3/4)

  y = x^2 +x +1

__

2. The zeros appear to be at x=1 and x=4, so the factored form of the equation is ...

  y = a(x -1)(x -4)

At x = 0, this is ...

  12 = a(-1)(-4)

  3 = a

So, the equation is ...

  y = 3x^2 -15x +12

The values at the two points of interest are ...

  y = 3(-2-1)(-2-4) = 54 . . . . point (-2, 54)

  y = 3(5 -1)(5 -4) = 12 . . . . point (5, 12)

__

3. Using the vertex shown, the vertex form of the equation is ...

  y = a(x -2)^2 +7

Filling in the point (-2, -1), we can find the value of a:

  -1 = a(-2-2)^2 +7

  -8 = 16a

  a = -8/16 = -1/2

So, the equation is ...

  y = -1/2x^2 +2x +5

For x=0, the y-value is 5. The y-intercept is (0, 5).

Answer:

9514 1404 393

y = x^2 +x +1; (-1/2, 3/4)

y = 3x^2 -15x +12; (-2, 54), (5, 12)

y = -1/2x^2 +2x +5; (0, 5)  

1. The graph looks like the vertex is at x = -1/2. So we can start with the equation y = (x +1/2)^2 +k. At x=0, this becomes 1 = 1/4 +k, or k = 3/4. The leading coefficient of 1 is consistent with the other points shown, so the equation is ...

 y = (x +1/2)^2 +3/4 . . . . vertex is (-1/2, 3/4)

 y = x^2 +x +1

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