Answer:
![f(x)=4(x-\frac{3}{2})^2-24](https://tex.z-dn.net/?f=f%28x%29%3D4%28x-%5Cfrac%7B3%7D%7B2%7D%29%5E2-24)
The minimum occurs at
.
Step-by-step explanation:
The
in
is the same
in
where:
and
.
So let's find
.
.
To find
, we must use the expression that
is and evaluate it for
, like so:
![4(\frac{3}{2})^2-12(\frac{3}{2})-15](https://tex.z-dn.net/?f=4%28%5Cfrac%7B3%7D%7B2%7D%29%5E2-12%28%5Cfrac%7B3%7D%7B2%7D%29-15)
![4(\frac{9}{4}-6(3)-15](https://tex.z-dn.net/?f=4%28%5Cfrac%7B9%7D%7B4%7D-6%283%29-15)
![9-18-15](https://tex.z-dn.net/?f=9-18-15)
![-9-15](https://tex.z-dn.net/?f=-9-15)
![-24](https://tex.z-dn.net/?f=-24)
So the vertex form is
.
-----------------Another way----------------------------
You could just complete the square.
I like to use the following to help me formulate the process:
.
Let's start. My formula requires coefficient of
to be 1 so factor out the 4 from the first two terms:
![f(x)=4x^2-12x-15](https://tex.z-dn.net/?f=f%28x%29%3D4x%5E2-12x-15)
![f(x)=4(x^2-3x)-15](https://tex.z-dn.net/?f=f%28x%29%3D4%28x%5E2-3x%29-15)
Now we are going to add the
to complete the square; whatever you add in you must also subtract out.
![f(x)=4(x^2-3x+(\frac{3}{2})^2)-15-4(\frac{3}{2})^2](https://tex.z-dn.net/?f=f%28x%29%3D4%28x%5E2-3x%2B%28%5Cfrac%7B3%7D%7B2%7D%29%5E2%29-15-4%28%5Cfrac%7B3%7D%7B2%7D%29%5E2)
![f(x)=4(x-\frac{3}{2})^2-15-4(\frac{9}{4})](https://tex.z-dn.net/?f=f%28x%29%3D4%28x-%5Cfrac%7B3%7D%7B2%7D%29%5E2-15-4%28%5Cfrac%7B9%7D%7B4%7D%29)
![f(x)=4(x-\frac{3}{2})^2-15-9](https://tex.z-dn.net/?f=f%28x%29%3D4%28x-%5Cfrac%7B3%7D%7B2%7D%29%5E2-15-9)
![f(x)=4(x-\frac{3}{2})^2-24](https://tex.z-dn.net/?f=f%28x%29%3D4%28x-%5Cfrac%7B3%7D%7B2%7D%29%5E2-24)
-----------------So anyways either way you choose....-----------------------
The minimum or maximum will occur at the vertex. Since
is positive the parabola is open up and therefore does have a minimum.
tells us the vertex is
.
So h is the x-coordinate of the vertex.
So the minimum occurs at
.