Hi I need this question ASAP
1. Solve the System y=x^2+2x-2 y=x+10
2. graph the function and identify the domain and range y=-2x^2
1. To solve the system y=x^2+2x-2 and y=x+10, we need to find the values for x and y where both equations are true. Since both equations are equal to y, we can set them equal to each other.
x^2+2x-2 = x+10
Now, solve the equation for x:
x^2+2x-2-x-10 = 0
x^2+x-12 = 0
Factor the quadratic equation:
(x+4)(x-3)=0
Now, set each factor equal to zero and solve for x:
x+4=0 -> x=-4
x-3=0 -> x=3
So, we have two values for x: -4 and 3. To find the values for y, plug these x values into either original equation. We'll use y=x+10.
For x=-4:
y=-4+10
y=6
For x=3:
y=3+10
y=13
So, we have two pairs of solutions: (-4,6) and (3,13).
2. To graph the function y=-2x^2, let's create a table of x and y values:
x y
-2 -8
-1 -2
0 0
1 -2
2 -8
The graph is a parabola opening downwards due to the negative coefficient of the x^2 term.
The domain of the function is all real numbers, since x can be any value:
Domain: (-∞, ∞)
The range of the function, however, is limited by the fact that the parabola opens downwards, and the highest point on the graph is at y=0 (the vertex). So the function has all y-values less than or equal to 0:
Range: (-∞, 0]