Respuesta :
Answer:
[tex]f(x)=4(x-\frac{3}{2})^2-24[/tex]
The minimum occurs at [tex]x=\frac{3}{2}[/tex].
Step-by-step explanation:
[tex]f(x)=4x^2-12x-15[/tex]
The [tex]a[/tex] in [tex]f(x)=ax^2+bx+c[/tex] is the same [tex]a[/tex] in [tex]f(x)=a(x-h)^2+k[/tex] where:
[tex]h=\frac{-b}{2a}[/tex] and [tex]k=f(\frac{-b}{2a})[/tex].
So let's find [tex]\frac{-b}{2a}=\frac{-(-12)}{2(4)}=\frac{12}{8}=\frac{3}{2}[/tex].
[tex]f(\frac{-b}{2a})=f(\frac{3}{2})[/tex].
To find [tex]f(\frac{3}{2})[/tex], we must use the expression that [tex]f[/tex] is and evaluate it for [tex]x=\frac{3}{2}[/tex], like so:
[tex]4(\frac{3}{2})^2-12(\frac{3}{2})-15[/tex]
[tex]4(\frac{9}{4}-6(3)-15[/tex]
[tex]9-18-15[/tex]
[tex]-9-15[/tex]
[tex]-24[/tex]
So the vertex form is [tex]f(x)=4(x-\frac{3}{2})^2-24[/tex].
-----------------Another way----------------------------
You could just complete the square.
I like to use the following to help me formulate the process:
[tex]x^2+dx+(\frac{d}{2})^2=(x+\frac{d}{2})^2[/tex].
Let's start. My formula requires coefficient of [tex]x^2[/tex] to be 1 so factor out the 4 from the first two terms:
[tex]f(x)=4x^2-12x-15[/tex]
[tex]f(x)=4(x^2-3x)-15[/tex]
Now we are going to add the [tex](\frac{d}{2})^2[/tex] to complete the square; whatever you add in you must also subtract out.
[tex]f(x)=4(x^2-3x+(\frac{3}{2})^2)-15-4(\frac{3}{2})^2[/tex]
[tex]f(x)=4(x-\frac{3}{2})^2-15-4(\frac{9}{4})[/tex]
[tex]f(x)=4(x-\frac{3}{2})^2-15-9[/tex]
[tex]f(x)=4(x-\frac{3}{2})^2-24[/tex]
-----------------So anyways either way you choose....-----------------------
The minimum or maximum will occur at the vertex. Since [tex]a[/tex] is positive the parabola is open up and therefore does have a minimum.
[tex]f(x)=a(x-h)^2+k[/tex] tells us the vertex is [tex](h,k)[/tex].
So h is the x-coordinate of the vertex.
So the minimum occurs at [tex]x=\frac{3}{2}[/tex].
