Example 1 Sketch the graph of each of the following parabolas.
Sumber
(a)
f (x) = 2(x + 3)2 − 8
(b)
g (x) = − (x – 2)2
− 1
(c)
h (x) = x2 + 4
Solution
Okay, in all of these we will simply go through the
process given above to find the needed points and the graph.
(a)
f (x) = 2(x + 3)2 – 8
First we need to find the vertex. We will need to be
careful with the signs however. Comparing our equation to the form above we see
that we must have h = −3
and k = −8 since
that is the only way to get the correct signs in our function. Therefore, the
vertex of this parabola is,
(−3,−8)
Now let’s find the y-intercept. This is nothing
more than a quick function evaluation.
f (0)
= 2 (0 + 3)2
− 8 = 2 (9)
− 8 =10 y − intercept : (0,10)
Next we need to find the x-intercepts. This
means we’ll need to solve an equation. However, before we do that we can
actually tell whether or not we’ll have any before we even start to solve the
equation.
In this case we have a = 2 which is positive and so we know that the parabola
opens up. Also the vertex is a point below the x-axis. So, we know that
the parabola will have at least a few points below the x-axis and it
will open up. Therefore, since once a parabola starts to open up it will
continue to open up eventually we will have to cross the x-axis. In
other words, there are x-intercepts for this parabola.
To find them we need to solve the following equation.
0 = 2 (x
+ 3)2 − 8
We solve equations like this back when we were solving quadratic equations so hopefully you remember how to do them.
2 (x + 3)2 = 8
(x
+ 3)2 = 4
x
+ 3 = ±√4 = ±2
x = – 3 ± 2 ⇒ x = – 1, x = – 5
The two x-intercepts are then,
(−5,0) and (−1,0)
Now, at this point we’ve got points on either side of the vertex so
we are officially done with finding the points. However, let’s talk a little
bit about how to find a second point using the y intercept and the axis
of symmetry since we will need to do that eventually.
First, notice that the y-intercept has an x coordinate
of 0 while the vertex has an x coordinate of - 3. This means that the y-intercept
is a distance of 3 to the right of the axis of symmetry since that will move
straight up from the vertex.
Now, the left part of the graph will be a mirror image of the right
part of the graph. So, since there is a point at y =10
that is a distance of 3 to the right of the axis of symmetry there must also be
a point at y =10 that is a distance of 3 to the
right of the axis of symmetry.
So, since the x coordinate of the vertex is -3 and this new
point is a distance of 3 to the left its x, coordinate must be -6. The
coordinates of this new point are then (−6,10). We
can verify this by evaluating the function at x = −6.
If we are correct we should get a value of 10. Let’s verify this.
f (−6) = 2 (−6 + 3)2
− 8 = 2 (− 3)2
− 8 = 2 (9)
− 8 =10
So, we were correct. Note that we usually don’t bother with the
verification of this point.
Note that we included the axis of symmetry in this
graph and typically we won’t. It was just included here since we were
discussing it earlier.
(b)
g (x) = − (x – 2)2
− 1
Okay with this one we won’t put in quite a much detail.
First let’s notice that a = −1 which is negative and so we know that this parabola will open
downward.
Next, by comparing our function to the general form we
see that the vertex of this parabola is (2,−1) . Again, be careful to get the signs correct here!
Now let’s get the y-intercept.
g (0)
= −( 0 – 2)2
− 1 = − (−2) − 1 = −4 −1 = −5
The y-intercept is then (0,−5).
Now, we know that the vertex starts out below the x-axis
and the parabola opens down. This means that there can’t possibly be x-intercepts
since the x axis is above the vertex and the parabola will always open
down. This means that there is no reason, in general, to go through the solving
process to find what won’t exist.
However, let’s do it anyway. This will show us what to
look for if we don’t catch right away that they won’t exist from the vertex and direction the
parabola opens. We’ll need to solve,
0 =
− (x – 2)2 – 1
(x – 2)2 = – 1
(x – 2)2 = – 1
x – 2 = ± i
x = 2 ± i
So, we got complex solutions. Complex solutions will always
indicate no x-intercepts.
Now, we do want points on either side of the vertex so we’ll use
the y-intercept and the axis of symmetry to get a second point. The y-intercept
is a distance of two to the left of the axis of symmetry and is at y = −5
and so there must be a second point at the same y value only a distance
of 2 to the right of the axis of symmetry. The coordinates of this point must
then be (4,−5).
(c)
h (x) = x2 + 4
This one is actually a fairly simple one to graph. We’ll first
notice that it will open upwards.
Now, the vertex is probably the point where most students run into
trouble here. Since we have x2 by itself this means that we
must have h = 0 and so the vertex is (0, 4) .
Note that this means there will not be any x-intercepts with
this parabola since the vertex is above the x-axis and the parabola
opens upwards.
Next, the y-intercept is,
h (0) = (0)2
+ 4 = 4 y − intercept
: 0, 4
The y-intercept is exactly the same as the vertex. This will
happen on occasion so we shouldn’t get too worried about it when that happens.
Although this will mean that we aren’t going to be able to use the y-intercept
to find a second point on the other side of the vertex this time. In fact, we
don’t even have a point yet that isn’t the vertex!
So, we’ll need to find a point on either side of the vertex. In
this case since the function isn’t too bad we’ll just plug in a couple of
points.
h (−2) = (−2)2 +
4 = 8 ⇒ (−2,8)
h (2) = (2)2 +4 =8 ⇒ (2,8)
Note that we could have gotten the second point here using the axis
of symmetry if we’d wanted to.
Example 2 Sketch the
graph of each of the following parabolas.
(a)
g (x) = 3x2
− 6x + 5
(b)
f (x) = −x2 + 8x
(c) f ( x) = x2 + 4x + 4
Solution
(a) For this parabola we’ve got a = 3, b = −6 and c = 5 .
Make sure that you’re careful with signs when identifying these values. So we
know that this parabola will open up since a is positive.
Here are the evaluations for the vertex.
The vertex is then (1, 2) .
Now at this point we also know that there won’t be any x-intercepts for this parabola since the vertex is above the x-axis and it opens upward.
The y-intercept is (0,5) and
using the axis of symmetry we know that (2,5) must
also be on the parabola.
(b) In
this case a = −1, b = 8
and c = 0 . From these we see that the parabola will open downward since a
is negative. Here are the vertex evaluations.
So, the vertex is (4,16) and
we also can see that this time there will be x-intercepts. In fact,
let’s go ahead and find them now.
0
= – x + 8x
0 = x ( – x + 8) ⇒ x
= 0, x = 8
So, the x-intercepts are (0,0) and (8,0) .
Notice that (0,0) is also the y-intercept. This
will happen on occasion so don’t get excited about it when it does.
(c)
In
this final part we have a = 1, b
= 4
and c = 4 . So, this parabola will open up.
Here are the vertex evaluations.
So, the vertex is (−2,0) .
Note that since the y coordinate of this point is zero it is also an x-intercept.
In fact it will be the only x-intercept for this graph. This makes sense
if we consider the fact that the vertex, in this case, is the lowest point on the
graph and so the graph simply can’t touch the x-axis anywhere else.
The fact that this parabola has only one x-intercept can be
verified by solving as we’ve done in the other examples to this point.
0
= x2 + 4x + 4
0 = (x + 2)2 ⇒ x
= – 2
Sure enough there is only one x-intercept. Note that this
will mean that we’re going to have to use the axis of symmetry to get a second
point from the y-intercept in this case.
Speaking of which, the y-intercept in this case is (0, 4) .
This means that the second point is (−4, 4) .
Example 3 Convert each of the following into the form f (x) = a (x – h)2
+ k .
(a)
f (x)
= 2x2 – 12 + 3
(b)
f (x) = −x2 +10x
− 1
Solution
Okay, as we pointed out above we are going to complete
the square here. However, it is a slightly different process than the other
times that we’ve seen it to this point.
(a) The thing that we’ve got to remember here is that we
must have a coefficient of 1 for the x2 term in order to
complete the square. So, to get that we will first factor the coefficient of
the x2 term out off the whole right side as follows.
Note that this will often put fractions into the
problem that is just something that we’ll need to be able to deal with. Also
note that if we’re lucky enough to have a coefficient of 1 on the x2
term we won’t have to do this step.
Now, this is where the process really starts differing
from what we’ve seen to this point. We still take one-half the coefficient of x
and square it. However, instead of adding this to both sides we do the
following with it.
We add and subtract this quantity inside the
parenthesis as show. Note that all we are really doing here is adding in zero
since 9-9=0! The order listed here is important. We MUST add first and then
subtract.
The next step is to factor the first three terms and
combine the last two as follows.
As a final step we multiply the 2 back through.
f (x)
= 2 (x −3)2 − 15
And there we go.
(b) Be careful here. We don’t have a coefficient of 1 on the x2 term,
we’ve got a coefficient of -1. So, the process is identical outside of that so
we won’t put in as much detail this time.
= – (x2
– 10x + 25 – 25
= – ((x – 5)2 – 24)
= – (x – 5)2
+ 24
Sumber
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