Question

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

Answer 1

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

Answer 2

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)

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

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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)

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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).