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Finding Intercepts, Domain, Range and Vertex of a Parabola - Concept 30,943 views
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
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
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
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.