Question

y = x^2+4x+3 if it had a concave down instead

y = -2(x - {blank})^2+{blank}

Answer 1

To rewrite the equation of the parabola \( y = x^2 + 4x + 3 \) in a format that represents a concave-down parabola, we first need to complete the square for the original equation, and then invert the sign of the quadratic coefficient.

### Step 1: Complete the square
The given quadratic is \( y = x^2 + 4x + 3 \).

1. Take the coefficient of \( x \) (which is 4), halve it (which gives 2), and square it (which gives 4).
2. Rewrite \( y \) as follows:

\[
y = (x^2 + 4x + 4) - 4 + 3
\]
\[
y = (x + 2)^2 - 1
\]

### Step 2: Rewrite as concave down
To make this concave down, we multiply the entire equation by -2:

\[
y = -2((x + 2)^2 - 1)
\]

Distributing -2 gives:

\[
y = -2(x + 2)^2 + 2
\]

### Final form
Thus, your parabola in the concave-down form is:

\[
y = -2(x + 2)^2 + 2
\]

Filling in the blanks in the requested format \( y = -2(x - \{blank}\})^2 + \{blank}\):

- For the first blank \((x + 2)\), it needs to be written as \( (x - (-2)) \), so the first blank is \(-2\).
- For the second blank \(\{blank\}\), we found it is \(2\).

Thus, we have:

\[
y = -2(x - (-2))^2 + 2
\]

So the final filled in is:
y = -2(x - (-2))^2 + 2.

Answer 2

sure?