Sure, let's convert the quadratic equation [tex]\( y = 8x^2 - 80x - 32 \)[/tex] to its vertex form by completing the square. Here are the steps:
### Step 1: Factor out the coefficient of [tex]\( x^2 \)[/tex] from the first two terms.
The given quadratic equation is:
[tex]\[ y = 8x^2 - 80x - 32 \][/tex]
First, factor out the coefficient 8 from the [tex]\( x^2 \)[/tex] and [tex]\( x \)[/tex] terms:
[tex]\[ y = 8(x^2 - 10x) - 32 \][/tex]
### Step 2: Complete the square inside the parentheses.
To complete the square, we need to find a constant that makes the expression inside the parentheses a perfect square trinomial. Take half of the coefficient of [tex]\( x \)[/tex] (which is -10), square it, and add and subtract it inside the parentheses:
[tex]\[ x^2 - 10x \][/tex]
Half of -10 is -5, and squaring it gives:
[tex]\[ (-5)^2 = 25 \][/tex]
We add and subtract 25 inside the parentheses:
[tex]\[ x^2 - 10x + 25 - 25 \][/tex]
So the expression becomes:
[tex]\[ y = 8(x^2 - 10x + 25 - 25) - 32 \][/tex]
[tex]\[ y = 8((x - 5)^2 - 25) - 32 \][/tex]
### Step 3: Distribute the coefficient 8.
Distribute the 8 through the terms in the parentheses:
[tex]\[ y = 8(x - 5)^2 - 8 \cdot 25 - 32 \][/tex]
[tex]\[ y = 8(x - 5)^2 - 200 - 32 \][/tex]
### Step 4: Simplify the constants.
Combine the constant terms:
[tex]\[ y = 8(x - 5)^2 - 232 \][/tex]
Thus, the vertex form of the given quadratic equation is:
[tex]\[ y = 8(x - 5)^2 - 232 \][/tex]
In this form, [tex]\( y = a(x - h)^2 + k \)[/tex], the vertex [tex]\((h, k)\)[/tex] of the parabola is [tex]\((5, -232)\)[/tex].
