Question

Write the equation of a parabola that has a vertex of (-3, -8), and passes through (-1, 0). Write it in the form f(x)=ax^2 + bx + c

Answer 1

To write the equation of a parabola in the form \( f(x) = ax^2 + bx + c \) given the vertex and a point through which the parabola passes, we can start with the vertex form of a parabola:

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

where \((h, k)\) is the vertex. Given that the vertex is \((-3, -8)\), we substitute \(h = -3\) and \(k = -8\):

\[
f(x) = a(x + 3)^2 - 8
\]

Next, we need to find the value of \(a\). We can use the point \((-1, 0)\) which the parabola passes through. Substituting \(x = -1\) and \(f(x) = 0\):

\[
0 = a(-1 + 3)^2 - 8
\]

Now simplify the equation:

\[
0 = a(2)^2 - 8
\]
\[
0 = 4a - 8
\]
\[
4a = 8
\]
\[
a = 2
\]

Now, substituting \(a\) back into the vertex form:

\[
f(x) = 2(x + 3)^2 - 8
\]

Next, we will convert this into standard form \(f(x) = ax^2 + bx + c\) by expanding:

\[
f(x) = 2(x^2 + 6x + 9) - 8
\]
\[
= 2x^2 + 12x + 18 - 8
\]
\[
= 2x^2 + 12x + 10
\]

Thus, the equation of the parabola in standard form is:

\[
f(x) = 2x^2 + 12x + 10
\]