Certainly! Let's work through the process of converting the quadratic expression [tex]\( y = 8x^2 + 32x + 17 \)[/tex] into its vertex form step-by-step.
### Step 1: Factor out the leading coefficient from the quadratic and linear terms
First, we factor out the leading coefficient (8) from the terms involving [tex]\( x \)[/tex]:
[tex]\[ y = 8(x^2 + 4x) + 17 \][/tex]
### Step 2: Form a perfect-square trinomial inside the parentheses
Next, we need to complete the square inside the parentheses. To do this, we take the coefficient of [tex]\( x \)[/tex] (which is 4), divide it by 2, and then square the result:
[tex]\[ \left( \frac{4}{2} \right)^2 = 2^2 = 4 \][/tex]
Now we add and subtract this value inside the parentheses:
[tex]\[ y = 8(x^2 + 4x + 4 - 4) + 17 \][/tex]
### Step 3: Simplify the expression inside the parentheses
The expression inside the parentheses can now be grouped as a perfect square trinomial:
[tex]\[ y = 8[(x^2 + 4x + 4) - 4] + 17 \][/tex]
[tex]\[ y = 8[(x + 2)^2 - 4] + 17 \][/tex]
### Step 4: Distribute the leading coefficient through the terms inside the parentheses
Next, distribute the 8 through the terms inside the brackets:
[tex]\[ y = 8(x + 2)^2 - 8 \cdot 4 + 17 \][/tex]
[tex]\[ y = 8(x + 2)^2 - 32 + 17 \][/tex]
### Step 5: Simplify the constant terms
Finally, combine the constant terms:
[tex]\[ y = 8(x + 2)^2 - 15 \][/tex]
### Conclusion
The vertex form of the quadratic expression [tex]\( y = 8x^2 + 32x + 17 \)[/tex] is:
[tex]\[ y = 8(x + 2)^2 - 15 \][/tex]
This represents the same quadratic equation in vertex form, where the vertex of the parabola is at [tex]\( (-2, -15) \)[/tex].
