Question

Rewrite the given quadratic function in the form f(x) = a(x - h)2 + k. Then, enter the x-coordinate of the minimum point of the function:

f(x)=4x^2-12x-15

Answer 1

`f(x)=4(x-(3)/(2))^2-24`

The minimum occurs at
`x=(3)/(2)`.

Explanation:

`f(x)=4x^2-12x-15`

The
`a` in
`f(x)=ax^2+bx+c` is the same
`a` in
`f(x)=a(x-h)^2+k` where:

`h=(-b)/(2a)` and
`k=f((-b)/(2a))`.

So let's find
`(-b)/(2a)=(-(-12))/(2(4))=(12)/(8)=(3)/(2)`.

`f((-b)/(2a))=f((3)/(2))`.

To find
`f((3)/(2))`, we must use the expression that
`f` is and evaluate it for
`x=(3)/(2)`, like so:

`4((3)/(2))^2-12((3)/(2))-15`

`4((9)/(4)-6(3)-15`

`9-18-15`

`-9-15`

`-24`

So the vertex form is
`f(x)=4(x-(3)/(2))^2-24`.

-----------------Another way----------------------------

You could just complete the square.

I like to use the following to help me formulate the process:

`x^2+dx+((d)/(2))^2=(x+(d)/(2))^2`.

Let's start. My formula requires coefficient of
`x^2` to be 1 so factor out the 4 from the first two terms:

`f(x)=4x^2-12x-15`

`f(x)=4(x^2-3x)-15`

Now we are going to add the
`((d)/(2))^2` to complete the square; whatever you add in you must also subtract out.

`f(x)=4(x^2-3x+((3)/(2))^2)-15-4((3)/(2))^2`

`f(x)=4(x-(3)/(2))^2-15-4((9)/(4))`

`f(x)=4(x-(3)/(2))^2-15-9`

`f(x)=4(x-(3)/(2))^2-24`

-----------------So anyways either way you choose....-----------------------

The minimum or maximum will occur at the vertex. Since
`a` is positive the parabola is open up and therefore does have a minimum.

`f(x)=a(x-h)^2+k` tells us the vertex is
`(h,k)`.

So h is the x-coordinate of the vertex.

So the minimum occurs at
`x=(3)/(2)`.