Name: Quadratic systems: a line and a parabola | Equations | Algebra 2 | Khan Academy
Uploaded: 2020-03-27T17:39:24.000Z
Duration: 6 min 12 s
Description: Thousands of YouTube videos with English-Chinese subtitles! Now you can learn to understand native speakers, expand your vocabulary, and improve your pronunciation...

- [Instructor] We're told the parabola given

Past simple

by y is equal to three x squared minus six x plus one

us is one intersection point is clearly identifiable

This is actually a screenshot from the exercise

But pause this video and see if you can answer

All right, so one intersection point is clearly identifiable

I see that one, and I see that one there.

This second one seems clearly identifiable

And what's interesting about these intersection points is,

because it's a point that sits on the graph

that it satisfies both of these equations,

that it's a solution to both of these equations.

So the other one is find the other intersection point.

So they want us to figure out this intersection point

to try to solve this system of equations,

and this is interesting because this is a system

of equations where one of the equations is not linear,

So let's see how we could go about doing that.

I have y is equal to three x squared minus six x plus one.

Well one way to tackle, and this is one way

to tackle any system of equations, is through substitution.

So if I can rewrite this linear equation as

into my first equation, into my quadratic one,

because this one is in red, and this one is the line

The easiest way to solve for y is to add x to both sides

and then all the rest of the stuff cancels out,

And so now we can substitute x minus one back

in for y, and so we get x minus one is equal

And on the right-hand side we get three x squared

Let's see, is there an obvious way to factor it?

that's equal to the product of three and two?

Three times two, and if this looks unfamiliar,

And actually, negative six and negative one work.

So what can do is I can rewrite this whole thing

for those of you are not familiar with this technique.

You could also use the quadratic formula.

so if I group these first two I can factor

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Quadratic systems: a line and a parabola | Equations | Algebra 2 | Khan Academy

stuff

obvious

figure

situation