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

Answer:

$f(x)=4(x-\frac{3}{2})^2-24$

The minimum occurs at $x=\frac{3}{2}$.

Step-by-step 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:

$h=\frac{-b}{2a}$ and $k=f(\frac{-b}{2a})$.

So let's find $\frac{-b}{2a}=\frac{-(-12)}{2(4)}=\frac{12}{8}=\frac{3}{2}$.

$f(\frac{-b}{2a})=f(\frac{3}{2})$.

To find $f(\frac{3}{2})$, we must use the expression that $f$ is and evaluate it for $x=\frac{3}{2}$, like so:

$4(\frac{3}{2})^2-12(\frac{3}{2})-15$

$4(\frac{9}{4}-6(3)-15$

$9-18-15$

$-9-15$

$-24$

So the vertex form is $f(x)=4(x-\frac{3}{2})^2-24$.

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

You could just complete the square.

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

$x^2+dx+(\frac{d}{2})^2=(x+\frac{d}{2})^2$.

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 $(\frac{d}{2})^2$ to complete the square; whatever you add in you must also subtract out.

$f(x)=4(x^2-3x+(\frac{3}{2})^2)-15-4(\frac{3}{2})^2$

$f(x)=4(x-\frac{3}{2})^2-15-4(\frac{9}{4})$

$f(x)=4(x-\frac{3}{2})^2-15-9$

$f(x)=4(x-\frac{3}{2})^2-24$

-----------------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 $x=\frac{3}{2}$.