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- anonymous

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- anonymous

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- anonymous

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).
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.
(x−4)2+(y+4)2=4
for the point (4,-2) is as follows:
(4−4)2+(−2+4)2=4
(0)2+(2)2=4
4=4

- anonymous

wiat...let me refiigure that line

- anonymous

The equation for the line is
y=14x−3
Now, let's show that the point (4,-2) makes this equation true just like it does for the circle.
Subbing in -2 for y and 4 for x, we get:
−2=14(4)−3
−2=1−3
−2=−2

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- anonymous

the linear equation is y=14x-3

- anonymous

|dw:1439475598238:dw|
here is the graph

- anonymous

dang...it didnt show...

- anonymous

ok here is the quadratic (x-4)^2+(y=4)^2=4

- anonymous

|dw:1439475897879:dw|

- anonymous

yes now the line is obviously your line and the circle is the quadratic

- anonymous

you plot it on the graph

- anonymous

sorry, i am not really good at showing my work. im only good at doing it

- OregonDuck

"@MeganChase went offline"

- anonymous

You guys were using a circle but the question asked for a quadratic (parabola). Is any shape acceptable?

- anonymous

ok. so scratch the stuff using this equation "(x-4)^2+(y=4)^2=4"
but keep the line y = 2x - 4 as the linear equation.
For a quadratic equation, keep it simple and just square the x to get y = 2x² - 4

- anonymous

graph y = 2x - 4 and y = 2x² - 4 here https://www.desmos.com/calculator

- anonymous

it is, but just so you know the solution to that isn't rational so you're going to have to deal with square roots/decimals

- anonymous

decimals*

- anonymous

ok. either way the process is the same. since they're both y = something, you can use substitution and drop the y's
y = 2x² - 4 and y = 2x - 4 are the equations so we can say
\[2x^2 - 4 = 2x - 4\]

- anonymous

Do you know how to solve that for x?

- anonymous

right so now subtract 2x from both sides
2x² - 2x = 0
Factor out 2x
2x(x - 1) = 0
Set each factor equal to 0 and solve
2x = 0
→ x = 0
x - 1 = 0
→ x = 1

- anonymous

you follow?

- anonymous

yes for part 2 it looks like all you have to do is graph the line and parabola and make sure you show the intersection points because that is the solution.
We still have to find the y-values for part 1 though. Just plug in 0 and 1 into either equation to get them.
For x = 0:
y = 2(0) - 4
For x = 1:
y = 2(1) - 4
And that will give the 2 intersection points/solutions

- anonymous

I don't know that it matters. They mean the same thing, just different notations. You can use y for both, or you can use function notation. If you use function notation and call the linear function f(x) then call the quadratic g(x). It doesn't matter as long as you're consistent throughout the problem

- anonymous

So your solutions to the system are (0, -4) and (1, -2). Which you can see from the graph
https://www.desmos.com/calculator/wdb2i4kk5d

- anonymous

you're welcome

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