Question

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

y=3(x+2)2+3

(1 point)
Responses

y=x2+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=3x2+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=3x2+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=3x2+12x+15
and (15,0)

Answer 1

To transform the quadratic equation \( y = 3(x + 2)^2 + 3 \) into standard form, we can first expand the equation:

1. Start by expanding \( (x + 2)^2 \):
\[
(x + 2)^2 = x^2 + 4x + 4
\]

2. Now substitute this back into the equation:
\[
y = 3(x^2 + 4x + 4) + 3
\]

3. Distribute the 3:
\[
y = 3x^2 + 12x + 12 + 3
\]
\[
y = 3x^2 + 12x + 15
\]

Now, we have the quadratic in standard form as \( y = 3x^2 + 12x + 15 \).

Next, to find the y-intercept, set \( x = 0 \):

\[
y = 3(0)^2 + 12(0) + 15 = 15
\]

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

Therefore, the final answer is:
**y = 3x^2 + 12x + 15 and (0, 15)**.