Answer:
See explanation and attachment
Step-by-step explanation:
a) To graph [tex]y=x^2-2x-1[/tex], we need to plot some few points.
When x=-2, [tex]y=(-2)^2-2(-2)-1=7[/tex] so we plot (-2,7).
When x=-1, [tex]y=(-1)^2-2(-1)-1=2[/tex] so we plot (-1,2)
When x=0, [tex]y=(0)^2-2(0)-1=-1[/tex] so we plot (0,-1)
When x=1, [tex]y=1^2-2(1)-1=-2[/tex] so we plot (1,-2)
When x=2, [tex]y=(2)^2-2(2)-1=-1[/tex] so we plot (2,-1)
We then draw a smooth curve through the points to obtain the curve in the attachment.
b) The graph intersected the x-axis (y=0) at x=-0.41 and x=2.41. These are the roots.
c) For the line y=2x-3
When x=0, y=2(0)-3=-3 so we plot (0,-3)
When x=1, y=2(1)-3=-1 So we plot (1,-1)
We draw a straight line through these two points to intersect the parabola as shown on the graph.
d) To solve
[tex]y=x^2-2x-1[/tex]
and
[tex]y=2x-3[/tex] simultaneously using the graph, we look for the point of intersection of the parabola and the straight line.
The solution is (0.59,-1.83) and (3.41,3.83)
