Question

PLS HELP ILL GIVE U POINTS :)

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

The said explanation is the answer.

Step-by-step explanation:

1) Graphing the first function (y = 2x-1). When the exponents of the variables are all 1, that is there are no square, cubes, etc. the equation is a line. Then it is enough to find two points to draw it. We can calculate two points assigning a number to x and calculating the corresponding number for y. For example, when x= 0 y= 2 times 0 -1 = -1. Then the line will pass from the point (0,-1). We need a second point, we can take x=1, then y = 2 times 1 -1 = 1. The line will pass from (1,1). We can mark the two points on the plane and draw the line from these two points.  

2) Graphing the second function (y = x^2-9). A quadratic equation is of the form: ax^2+bx+c= 0. We have: y= f(x) -9. Set y=0. We have the quadratic equation: x^2-9=0. Using the algebraic identity: a^2-b^2=(a+b)(a-b). We can rewrite x^2-9=0 as x^2-3^2=0: [(x+3)(x-3)=0] - [(x+3)=0,(x-3)=0] - [x= -3, x= 3]. Hence, there are two solutions for x. So we have two x-intercepts: (-3,0) and (3,0). To find the y-intercept, set x=0: [(y = (0)^2-9=0)] - [(y=-9)]. Hence the y-intercept: (0,-9)

In conclusion, this system of equations at most has two real solutions, since two points of both functions intersect each other.

Here's the graph:

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