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].
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