1. First we are going to find the vertex of the quadratic function
![f(x)=2x^2+8x+1](/tpl/images/0294/5181/42198.png)
. To do it, we are going to use the vertex formula. For a quadratic function of the form
![f(x)=ax^2+bx +c](/tpl/images/0294/5181/19fff.png)
, its vertex
![(h,k)](/tpl/images/0294/5181/c3e72.png)
is given by the formula
![h= \frac{-b}{2a}](/tpl/images/0294/5181/b5eed.png)
;
![k=f(h)](/tpl/images/0294/5181/f8ec9.png)
.
We can infer from our problem that
![a=2](/tpl/images/0294/5181/b272a.png)
and
![b=8](/tpl/images/0294/5181/1b6fe.png)
, sol lets replace the values in our formula:
![h= \frac{-8}{2(2)}](/tpl/images/0294/5181/20fe7.png)
![h= \frac{-8}{4}](/tpl/images/0294/5181/23c77.png)
![h=-2](/tpl/images/0294/5181/ee8ee.png)
Now, to find
![k](/tpl/images/0294/5181/cc0ac.png)
, we are going to evaluate the function at
![h](/tpl/images/0294/5181/8a40e.png)
. In other words, we are going to replace
![x](/tpl/images/0294/5181/a0e3f.png)
with -2 in the function:
![k=f(-2)=2(-2)^2+8(-2)+1](/tpl/images/0294/5181/91eec.png)
![k=f(-2)=2(4)-16+1](/tpl/images/0294/5181/c7270.png)
![k=f(-2)=8-16+1](/tpl/images/0294/5181/37829.png)
![k=f(-2)=-7](/tpl/images/0294/5181/be177.png)
![k=-7](/tpl/images/0294/5181/9a8a3.png)
So, our first point, the vertex
![(h,k)](/tpl/images/0294/5181/c3e72.png)
of the parabola, is the point
![(-2,-7)](/tpl/images/0294/5181/05428.png)
.
To find our second point, we are going to find the y-intercept of the parabola. To do it we are going to evaluate the function at zero; in other words, we are going to replace
![x](/tpl/images/0294/5181/a0e3f.png)
with 0:
![f(x)=2x^2+8x+1](/tpl/images/0294/5181/42198.png)
![f(0)=2(0)^2+(0)x+1](/tpl/images/0294/5181/2c4e3.png)
![f(0)=1](/tpl/images/0294/5181/03db2.png)
So, our second point, the y-intercept of the parabola, is the point (0,1)
We can conclude that using the vertex (-2,-7) and a second point we can graph
![f(x)=2x^2+8x+1](/tpl/images/0294/5181/42198.png)
as shown in picture 1.
2. The vertex form of a quadratic function is given by the formula:
![f(x)=a(x-h)^2+k](/tpl/images/0294/5181/835d7.png)
where
![(h,k)](/tpl/images/0294/5181/c3e72.png)
is the vertex of the parabola.
We know from our previous point how to find the vertex of a parabola.
![h= \frac{-b}{2a}](/tpl/images/0294/5181/b5eed.png)
and
![k=f(h)](/tpl/images/0294/5181/f8ec9.png)
, so lets find the vertex of the parabola
![f(x)=x^2+6x+13](/tpl/images/0294/5181/1f061.png)
.
![a=1](/tpl/images/0294/5181/f0724.png)
![b=6](/tpl/images/0294/5181/b4df7.png)
![h= \frac{-6}{2(1)}](/tpl/images/0294/5181/0fc0d.png)
![h=-3](/tpl/images/0294/5181/f22a6.png)
![k=f(-3)=(-3)^2+6(-3)+13](/tpl/images/0294/5181/93d9f.png)
![k=4](/tpl/images/0294/5181/af626.png)
Now we can use our formula to convert the quadratic function to vertex form:
![f(x)=a(x-h)^2+k](/tpl/images/0294/5181/835d7.png)
![f(x)=1(x-(-3))^2+4](/tpl/images/0294/5181/380f3.png)
![f(x)=(x+3)^2+4](/tpl/images/0294/5181/188c1.png)
We can conclude that the vertex form of the quadratic function is
![f(x)=(x+3)^2+4](/tpl/images/0294/5181/188c1.png)
.
3. Remember that the x-intercepts of a quadratic function are the zeros of the function. To find the zeros of a quadratic function, we just need to set the function equal to zero (replace
![f(x)](/tpl/images/0294/5181/3abe1.png)
with zero) and solve for
![x](/tpl/images/0294/5181/a0e3f.png)
.
![f(x)=x^2+4x-60](/tpl/images/0294/5181/f69ea.png)
![0=x^2+4x-60](/tpl/images/0294/5181/8d439.png)
![x^2+4x-60=0](/tpl/images/0294/5181/d9fbf.png)
To solve for
![x](/tpl/images/0294/5181/a0e3f.png)
, we need to factor our quadratic first. To do it, we are going to find two numbers that not only add up to be equal 4 but also multiply to be equal -60; those numbers are -6 and 10.
![(x-6)(x+10)=0](/tpl/images/0294/5181/9767e.png)
Now, to find the zeros, we just need to set each factor equal to zero and solve for
![x](/tpl/images/0294/5181/a0e3f.png)
.
![x-6=0](/tpl/images/0294/5181/7f6ae.png)
and
![x+10=0](/tpl/images/0294/5181/cd6a2.png)
![x=6](/tpl/images/0294/5181/bdba1.png)
and
![x=-10](/tpl/images/0294/5181/a0cb2.png)
We can conclude that the x-intercepts of the quadratic function
![f(x)=x^2+4x-60](/tpl/images/0294/5181/f69ea.png)
are the points (0,6) and (0,-10).
4. To solve this, we are going to use function transformations and/or a graphic utility.
Function transformations.
- Translations:
We can move the graph of the function up or down by adding a constant
![c](/tpl/images/0294/5181/7e8ab.png)
to the y-value. If
![c\ \textgreater \ 0](/tpl/images/0294/5181/abcf9.png)
, the graph moves up; if
![c\ \textless \ 0](/tpl/images/0294/5181/82be1.png)
, the graph moves down.
- We can move the graph of the function left or right by adding a constant
![c](/tpl/images/0294/5181/7e8ab.png)
to the x-value. If
![c\ \textgreater \ 0](/tpl/images/0294/5181/abcf9.png)
, the graph moves left; if
![c\ \textless \ 0](/tpl/images/0294/5181/82be1.png)
, the graph moves right.
- Stretch and compression:
We can stretch or compress in the y-direction by multiplying the function by a constant
![c](/tpl/images/0294/5181/7e8ab.png)
. If
![c\ \textgreater \ 1](/tpl/images/0294/5181/cde65.png)
, we compress the graph of the function in the y-direction; if
![0\ \textless \ c\ \textless \ 1](/tpl/images/0294/5181/bd4f1.png)
, we stretch the graph of the function in the y-direction.
We can stretch or compress in the x-direction by multiplying
![x](/tpl/images/0294/5181/a0e3f.png)
by a constant
![c](/tpl/images/0294/5181/7e8ab.png)
. If
![c\ \textgreater \ 1](/tpl/images/0294/5181/cde65.png)
, we compress the graph of the function in the x-direction; if
![0\ \textless \ c\ \textless \ 1](/tpl/images/0294/5181/bd4f1.png)
, we stretch the graph of the function in the x-direction.
a. The
![c](/tpl/images/0294/5181/7e8ab.png)
value of
![f(x)](/tpl/images/0294/5181/3abe1.png)
is 2; the
![c](/tpl/images/0294/5181/7e8ab.png)
value of
![g(x)](/tpl/images/0294/5181/1061e.png)
is -3. Since
![c](/tpl/images/0294/5181/7e8ab.png)
is added to the whole function (y-value), we have an up/down translation. To find the translation we are going to ask ourselves how much should we subtract to 2 to get -3?
![c+2=-3](/tpl/images/0294/5181/ead65.png)
![c=-5](/tpl/images/0294/5181/060d0.png)
Since
![c\ \textless \ 0](/tpl/images/0294/5181/82be1.png)
, we can conclude that the correct answer is: It is translated down 5 units.
b. Using a graphing utility to plot both functions (picture 2), we realize that
![g(x)](/tpl/images/0294/5181/1061e.png)
is 1 unit to the left of
![f(x)](/tpl/images/0294/5181/3abe1.png)
We can conclude that the correct answer is: It is translated left 1 unit.
c. Here we have that
![g(x)](/tpl/images/0294/5181/1061e.png)
is
![f(x)](/tpl/images/0294/5181/3abe1.png)
multiplied by the constant term 2. Remember that We can stretch or compress in the y-direction (vertically) by multiplying the function by a constant
![c](/tpl/images/0294/5181/7e8ab.png)
.
Since
![c\ \textgreater \ 0](/tpl/images/0294/5181/abcf9.png)
, we can conclude that the correct answer is: It is stretched vertically by a factor of 2.
![Can someone me with this? i still have 2 more pages to do and i'm stressed out of my mind i hones](/tpl/images/0294/5181/86f11.jpg)
![Can someone me with this? i still have 2 more pages to do and i'm stressed out of my mind i hones](/tpl/images/0294/5181/16e16.jpg)