# Finding Intercepts, Domain, Range and Vertex of a Parabola - Concept

When graphing and describing the characteristics of a parabola, it is important to know several key pieces of information. The **parabola intercepts** describe where the parabola intersects the x-axis and the y-axis while the vertex of a parabola is the highest (or lowest) point of the parabola. Knowing the domain and range of a parabola is also helpful when graphing.

Finding the intercepts domain and range

and another way to find the vertex

a parabola.

So whenever we're finding X intercepts,

what we want to do is let Y is equal

to 0. So we let Y equal 0.

What we end up with is 0 is equal to X squared

minus 5X plus 6. So what we

end up with is a quadratic equation,

which we have the tools to solve.

We can either factor, complete the square,

quadratic formula, a number of different

ways to solve this out. This particular example

factors quite easily. We end up with X minus

2. X minus 3. So we know that our intercepts on R2

and 3. That's easy enough for our

X intercepts.

For our Y intercepts, what we want to

do is let X equal 0. When we let

X equals 0 our first term disappears,

our second term disappears, and

we're just left with our constant term

which in this case is just going

to be 6.

Finding the vertex.

So two ways of finding the vertex.

We can either complete the square, which

tends to be pretty involved, or there

is a little shortcut which is negative B

over 2A is equal to the X coordinate.

of the vertex.

So all we have to do is go to our equation.

Remember the coefficient on X squared

is A. Coefficient on X is B and

the constant term is C. So negative

B over 2A in this case is just

going to be negative, negative 5, 5 over

2 times A which is just 2 times

1. So we find out that our X coordinate

is 5 over 2. Okay.

In order to find the other part of our vertex,

we just found the X coordinate.

We still have to find the Y. All we have

to do is plug in the X coordinate

into this equation.

This particular example isn't the nicest,

because we're dealing with a fraction

but it still doesn't matter all we have

to do is plug in two and a half.

We end up with two and a half squared minus

5 times 2.5 plus 6 and we end up

with negative 1.25. So by plugging in that

negative B over 2A into the equation we end up with our

Y coordinate of our vertici.

To find the domain, domain is value of

X that we have to put in, there's no

restrictions. We're not dividing by 0.

There's no square roots.

So X can actually be whatever it wants.

So we can call it all reals,

negative infinity to

positive infinity, different ways of saying

the same thing.

The last thing we're looking at is the range

of the Y values. This is going to take a little

bit of piecing together.

So we know we have a parabola.

Coefficient on the X squared term is 1,

which means our parabola is going to

be facing upwards.

So we know we have an upward

facing parabola.

And we know that our vertex is

at a point 5/2s and 1/4.

So the Y coordinate of our vertex is going to be the lowest point

upward facing parabola.

Y coordinate at the bottom.

So what we end up is our range being from

negative 1/4 to infinity.

Our range actually hits that point.

That point is actually

a point on the curve.

So we then can include negative

1/4 and we're going up.

If this was a negative coefficient on the

X squared term, I know that the parabola

would be going down.

So I would know that I'm going from negative

infinity all the way up to that

Y coordinate of the vertex.

So X intercept and Y intercept are similar

to finding any other X intercept,

Y intercept, let the other coordinate

equal 0 and solve.

Little trick for finding the vertex.

Negative V 2 over A plug it in to find

the Y coordinate and domain and range

and domain is going to be all reallies

for parabola, range just consider

the vertex and if the graph

is going up or down.

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