Question

Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s).
Consider the equation below.

y=3x^2+30x+71

Use completing the square to rewrite the given equation and reveal the extreme value.

y=3(x+___)^2+___



The extreme value of the equation is at ( ___ , ___).

Answer 1

To complete the square for the equation \( y = 3x^2 + 30x + 71 \), follow these steps:

1. Factor out the coefficient of \( x^2 \) (which is 3) from the first two terms:
\[
y = 3(x^2 + 10x) + 71
\]

2. Complete the square inside the parentheses. To do this, take the coefficient of \( x \) (which is 10), divide it by 2 (giving 5), and then square it (getting 25). Add and subtract this value inside the parentheses:
\[
y = 3(x^2 + 10x + 25 - 25) + 71
\]
\[
y = 3((x + 5)^2 - 25) + 71
\]

3. Distribute the 3:
\[
y = 3(x + 5)^2 - 75 + 71
\]
\[
y = 3(x + 5)^2 - 4
\]

Now we can fill in the boxes:

- \( y = 3(x + 5)^2 - 4 \)

So:
- The complete square form is \( y = 3(x + 5)^2 - 4 \), giving:
- First blank: \( 5 \)
- Second blank: \( -4 \)

The extreme value occurs at the vertex of the parabola. The vertex (extreme value) is at the point \( (h, k) \), where \( h = -5 \) and \( k = -4 \).

Thus, the extreme value of the equation is at:
- \( (-5, -4) \)

In conclusion, the final answers are:
- \( y = 3(x + 5)^2 - 4 \)
- Extreme value at:
- First blank: \( -5 \)
- Second blank: \( -4 \)

So, the answers to fill in are:
- \( 5 \), \( -4 \), \( -5 \), \( -4 \).