To graph the parabola, we first need to identify a, b, and c.
y=x^2-2x-1 ⇔ y= 1x^2+( - 2)x+( -1)
For this equation we have that a= 1, b= - 2, and c= -1. Now, we can find the vertex using its formula. To do this, we will need to think of y as a function of x, y=f(x).
Vertex of a Parabola: ( - b/2 a,f(- b/2 a) )Let's find the x-coordinate of the vertex.
The y-coordinate of the vertex is -2. Thus, the vertex is at the point (1,- 2). With this, we also know that the axis of symmetry of the parabola is the line x=1. Next, let's find two more points on the curve, one on each side of the axis of symmetry.
x
x^2-2x-1
y=x^2-2x-1
^2-2( )-1
-1
2
2^2-2(2)-1
-1
Both ( ,-1) and (2,-1) are on the graph. Let's form the parabola by connecting these points and the vertex with a smooth curve.
Graphing the Line
Let's now graph the linear function on the same coordinate plane. For a linear equation written in slope-intercept form, we can identify its slope m and y-intercept b.
y=- x-1 ⇔ y=-1x+( -1)
Both, the slope of the line and the y-intercept are -1.
Finding the Solutions
Finally, let's try to identify the coordinates of the points of intersection of the parabola and the line.
We can see that the points of intersection occur at (0,-1) and (1,-2). These are the solutions to the system.
