Can someone please help me with these questions! Please explain to me how to do it as well! Thank You!

  (2)  (-2, 4), (3, 9)

  (3)  (-2, 4), (1, 1)

Step-by-step explanation:

My favorite way to answer questions like this is to use a graphing calculator to find the points of intersection. The attached shows the points of interest.

If you want to solve this by hand, equate the expressions for y, then solve the resulting quadratic using any of the methods you know. Once you find the x-values, put those in either equation to find the y-value.

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(2) x^2 = y = x+6

  x^2 -x -6 = 0 . . . . put in standard form

  (x -3)(x +2) = 0 . . . . factor

  x = 3 or -2 . . . . . find the x-values that make the factors zero

  y = x^2 = 9 or 4

The points are (3, 9) and (-2, 4).

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(3) x^2 = -x +2

  x^2 +x -2 = 0 . . . . standard form

  (x +2)(x -1) = 0 . . . factored

  x = -2 or 1 . . . . . . .zeros

  y = x^2 = 4 or 1

The points are (-2, 4) and (1, 1).

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Comment on factoring

When you consider the factored form and its expanded equivalent, ...

  (x +a)(x +b) = x^2 +(a+b)x +ab

you see that the coefficient of the linear term (a+b) is the sum of factors of the constant term (ab). Your knowledge of multiplication tables will often help you factor equations easily. You need to pay attention to signs.

In problem 2, we want factors of -6 that have a sum of -1:

  -6 = (-1)(6) = (-2)(3) = (-3)(2) = (-6)(1)

These factor pairs have sums of 5, 1, -1, -5. The pair whose sum is -1 is the one we're looking for: -3 and 2. So, the factors in problem 2 are (x-3)(x+2).

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In problem 3, we want factors of -2 that have a sum of +1:

  -2 = (-1)(2) = (-2)(1)

These factor pairs have sums of 1, -1. So, the second pair is the one of interest: -2 and 1. That makes the factors be (x-2)(x+1).

B, Supplementary

Step-by-step explanation:

Two angles are supplementary when both of them add up to ONLY 180.