Question

Create a system of equations that includes one linear equation and one quadratic equation.

Part 1. Show all work to solving your system of equations algebraically.

Part 2. Graph your system of equations, and show the solution graphically to verify your solution.

Answer 1

Actually if I were doing this I would start by working backwards. I would draw the graphs in the same plane and then find their equations from the points and intercepts and whatnot. Like this, for example:

Answer 2

I'm really bad at this, sorry :/

Answer 3

That looks kinda crazy, I know, but it's because of the lack of accuracy and the fact I have no graph paper. That is a circle that has an equation of \[(x-4)^{2}+(y+4)^{2}=4\]and a line that intersects (or touches) the circle at the point (4,-2) with a slope of 2. The line has a y intercept of -4. So the equation for the line is\[y = 2x-4\]and the point in common, aka the solution, to that system is (4,-2).

Answer 4

The way you would do the algebra part is to show that the point (4,-2) works in both equations to make them both true.

Answer 5

Like this for the circle:

Answer 6

so what would i write?

Answer 7

\[(x-4)^{2}+(y+4)^{2}=4\]for the point (4,-2) is as follows:

Answer 8

\[(4-4)^{2}+(-2+4)^{2}=4\]\[(0)^{2}+(2)^{2}=4\]\[4=4\]

Answer 9

For the line using the same point (4,-2):

Answer 10

wait...that line doesn't work out. Hold on...let me refigure the line's equation. Don't go anywhere...

Answer 11

ok

Answer 12

The equation for the line is\[y=\frac{ 1 }{4 }x-3\]

Answer 13

Now, let's show that the point (4,-2) makes this equation true just like it does for the circle.

Answer 14

Subbing in -2 for y and 4 for x, we get:\[-2=\frac{ 1 }{ 4 }(4)-3\]\[-2=1-3\]\[-2=-2\]

Answer 15

Those are your equations and that is how you show your work algebraically.

Answer 16

So this is actually the graph you want to show:

Answer 17

Dude you are a huge help.