Complete the statements below that show [tex]$y=8x^2+32x+17$[/tex] being converted to vertex form.

1. Factor out the leading coefficient:
[tex]y=8(x^2+4x)+17[/tex]

2. Form a perfect-square trinomial:
[tex]y=8(x^2+4x+\square)+17-\square[/tex]

Fill in the blanks to complete the square and rewrite the equation in vertex form.



Answer :

Sure, let's proceed step-by-step to convert the quadratic equation [tex]\( y = 8x^2 + 32x + 17 \)[/tex] into vertex form.

1. Factor out the leading coefficient:
[tex]\[ y = 8(x^2 + 4x) + 17 \][/tex]

2. Form a perfect-square trinomial:

To complete the square inside the parenthesis, we need to add and subtract the square of half the coefficient of [tex]\( x \)[/tex]. The coefficient of [tex]\( x \)[/tex] in [tex]\( x^2 + 4x \)[/tex] is 4. Half of 4 is 2, and squared, it is [tex]\( 2^2 = 4 \)[/tex].

Add and subtract this square inside the parenthesis:
[tex]\[ y = 8\left(x^2 + 4x + 4 - 4\right) + 17 \][/tex]

Simplify:
[tex]\[ y = 8\left((x^2 + 4x + 4) - 4\right) + 17 \][/tex]

Factor the trinomial into a perfect square and distribute the 8:
[tex]\[ y = 8(x + 2)^2 - 8(4) + 17 \][/tex]
Which simplifies to:
[tex]\[ y = 8(x + 2)^2 - 32 + 17 \][/tex]

Therefore, we get:
[tex]\[ y = 8(x + 2)^2 - 15 \][/tex]

Now the quadratic equation [tex]\( y = 8x^2 + 32x + 17 \)[/tex] is converted to the vertex form [tex]\( y = 8(x + 2)^2 - 15 \)[/tex].

So, completing the statements:
[tex]\[ y = 8 \left( x^2 + 4x + 4 \right) - 32 + 17 \][/tex]

And the perfect-square trinomial itself is [tex]\( 8(x + 2)^2 \)[/tex].