Hi I need this question ASAP!

1.Solve the System y=x^2+2x-1 y=x+10

2. y=-2x^2 domain and range on a graph

1 answer

1. To solve the system y = x^2 + 2x - 1 and y = x + 10, set the two equations equal to each other:

x^2 + 2x - 1 = x + 10

Now, subtract x and 10 from both sides of the equation:

x^2 + x - 11 = 0

Factor the equation:

(x - 4)(x + 3) = 0

Solve for x:

x = 4 or x = -3

Now, plug these x values back into either original equation to find their corresponding y-values. Using the second equation, y = x + 10:

When x = 4, y = 4 + 10 = 14
When x = -3, y = -3 + 10 = 7

So the solutions to the system are:

(x, y) = (4, 14) and (x, y) = (-3, 7)

2. For the equation y = -2x^2, first find the vertex of the parabola. Since it's a quadratic equation in the form y = ax^2 + bx + c, where a = -2, b = 0, and c = 0, the vertex is located at the point (h, k) where:

h = -b / (2a) = -0 / (2 * -2) = 0
k = -2 * h^2 = -2 * 0^2 = 0

Thus, the vertex is at the point (0, 0). Since the parabola opens downward (because a = -2, which is negative), the highest point is at the vertex.

Domain: The domain is all real numbers, since the function is quadratic and has no restrictions on the x-values. So, the domain is (-∞, ∞).

Range: The highest value of y occurs at the vertex, which is at y = 0. Since the parabola opens downward, the y-values will be all values less than or equal to 0. So the range is (-∞, 0].