Question

Graph is open downwards vertex is (-2,5) and crosses the y at 1
Write in standard form

Answer 1

To write the equation of a downward-opening parabola in standard form, we use the vertex form of a parabola:

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

where \((h, k)\) is the vertex. In your case, the vertex is \((-2, 5)\). Thus, the equation starts as:

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

Next, we need to find the value of \(a\). Since the parabola crosses the y-axis at \(y = 1\), we can use the point where \(x = 0\) to find \(a\).

Substituting \(x = 0\) and \(y = 1\) into the equation:

\[
1 = a(0 + 2)^2 + 5
\]

This simplifies to:

\[
1 = 4a + 5
\]

Subtract 5 from both sides:

\[
1 - 5 = 4a
\]
\[
-4 = 4a
\]

Now, divide both sides by 4:

\[
a = -1
\]

Now that we have found \(a\), we can write the equation of the parabola in standard form:

\[
y = -1(x + 2)^2 + 5
\]

If you prefer to convert this into the general form \(y = ax^2 + bx + c\), we can expand it:

\[
y = -1(x^2 + 4x + 4) + 5
\]
\[
y = -x^2 - 4x - 4 + 5
\]
\[
y = -x^2 - 4x + 1
\]

So the equation of the parabola in standard form is:

\[
y = -1(x + 2)^2 + 5
\]

And in general form, it is:

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