Question

Help parabola?

Answer 1

what do i do?

Answer 2

This is a bit involved to describe by typing, but I'll try.

First thing you need to do is to plot \(y = x^2+1\). Do you know how to do that? Make a table of x values and y values, two columns:

x y
-4
-3
-2
-1
0
1
2
3
4

for each value of x, compute x^2+1 and write it down in the y column of the same row.
When you are done, plot those points on your graph paper and run a smooth curve through them.

Answer 3

Have you made the graph yet?

Answer 4

yeah

Answer 5

Okay. that's the first part done. Now, the next part is to use the graph to find the solutions to \(x^2+1=2\). Draw a line across the graph that represents y = 2. The points where that crosses the graph are the spots where \(x^2+1=2\). Do you see why that is?

Answer 6

yeah i understand the first one.
Its the second i dont know

Answer 7

For the solution of \[x^2-x+1=6\]we need to draw a different line. This line is the line \(y = x+6\). It will cross the parabola at the spots where \(x^2 - x + 1 = 6\). If you do a little bit of algebra, you can see that
\[x^2-x+1=6\]\[x^2+1=6+x\]\[x^2+1\]is the parabola that have plotted, and \[y=6+x\]is the additional line we draw.

For the third part, we'll do a similar procedure to solve \(x^2+x-7=0\):
\[x^2+x-7 = 0\][x^2-7=-x\]\[x^2+1-8 = -x\]\[x^2+1=8-x\]So for that one you need to draw in the line \(y = 8-x\) and the intersection points with the parabola are the solutions to \(x^2+x-7=0\)

Answer 8

oops, I see I missed a character in there, that should been \[x^2-7=-x\] instead of [x^2-7=-x\]

Answer 9

When you're all done, the graph ought to look this this: