Question

Transform the quadratic equation into standard form and find the y-intercept.

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

(1 point)
Responses

y=x^2+4x+5
and (0,15)
y is equal to x squared plus 4 x plus 5 and open paren 0 comma 15 close paren

y=3x^2+12x+5
and (0,15)
y is equal to 3 x squared plus 12 x plus 5 and open paren 0 comma 15 close paren

y=3x^2+12x+15
and (0,15)
y is equal to 3 x squared plus 12 x plus 15 and open paren 0 comma 15 close paren

y=3x^2+12x+15
and (15,0)

Answer 1

To transform the given quadratic equation \( y = 3(x + 2)^2 + 3 \) into standard form and find the y-intercept, we follow these steps:

1. **Expand the equation**:

Start with the expression:
\[
y = 3(x + 2)^2 + 3
\]

First, expand \( (x + 2)^2 \):
\[
(x + 2)^2 = x^2 + 4x + 4
\]

Now, substitute back:
\[
y = 3(x^2 + 4x + 4) + 3
\]
\[
y = 3x^2 + 12x + 12 + 3
\]
\[
y = 3x^2 + 12x + 15
\]

2. **Identify the y-intercept**:

The y-intercept occurs when \( x = 0 \). Substitute \( x = 0 \):
\[
y = 3(0)^2 + 12(0) + 15 = 15
\]

Thus, the y-intercept is \( (0, 15) \).

So, the standard form is \( y = 3x^2 + 12x + 15 \) and the y-intercept is \( (0, 15) \).

Thus, the correct response is:
**y = 3x^2 + 12x + 15 and (0, 15)**.