Question

convert the quadratic y+(x+4)^2 - 5 from vortex form to standard form

Answer 1

To convert the quadratic \( y + (x + 4)^2 - 5 \) from vertex form to standard form, we first start with the given equation in vertex form:

\[
y + (x + 4)^2 - 5 = 0
\]

We can simplify this to:

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

Now, let’s express it in standard form \( y = ax^2 + bx + c \).

1. Expand the squared term:
\[
y = -((x + 4)(x + 4)) + 5
\]

2. Use the FOIL method (First, Outer, Inner, Last):
\[
(x + 4)(x + 4) = x^2 + 8x + 16
\]

3. Substitute it back into the equation:
\[
y = -(x^2 + 8x + 16) + 5
\]

4. Distribute the negative sign:
\[
y = -x^2 - 8x - 16 + 5
\]

5. Combine like terms:
\[
y = -x^2 - 8x - 11
\]

Thus, the quadratic in standard form is:

\[
y = -x^2 - 8x - 11
\]