To find the vertex form of the quadratic function \( f(x) \) that has roots -2 and 6 and passes through the point (1, 15), we follow these steps:
1. Start with the factored form of the quadratic equation:
Since the quadratic function needs to have roots at -2 and 6, it can be written in factored form as:
[tex]\[
f(x) = a(x + 2)(x - 6)
\][/tex]
Here, \(a\) is a constant that we need to determine.
2. Use the point (1, 15) to solve for \(a\):
The function passes through the point (1, 15), which means when \(x = 1\), \(f(x) = 15\). Substitute \(x = 1\) and \(f(x) = 15\) into the equation:
[tex]\[
15 = a(1 + 2)(1 - 6)
\][/tex]
Simplify inside the parentheses:
[tex]\[
15 = a(3)(-5)
\][/tex]
[tex]\[
15 = a \cdot -15
\][/tex]
Solve for \(a\):
[tex]\[
15 = -15a
\][/tex]
[tex]\[
a = -1
\][/tex]
3. Write the quadratic function with the determined \(a\) value:
Now we know that \(a = -1\), so:
[tex]\[
f(x) = -1(x + 2)(x - 6)
\][/tex]
4. Expand the function to convert it to standard form:
Expand the product:
[tex]\[
f(x) = -1(x^2 - 4x - 12)
\][/tex]
Distribute \(-1\):
[tex]\[
f(x) = -x^2 + 4x + 12
\][/tex]
5. Convert the function to vertex form:
To convert \(f(x) = -x^2 + 4x + 12\) to vertex form, we complete the square.
- First, factor out \(-1\) from the quadratic and linear terms:
[tex]\[
f(x) = -(x^2 - 4x) + 12
\][/tex]
- To complete the square inside the parentheses, add and subtract \(\left(\frac{4}{2}\right)^2 = 4\):
[tex]\[
f(x) = -(x^2 - 4x + 4 - 4) + 12
\][/tex]
[tex]\[
f(x) = -((x^2 - 4x + 4) - 4) + 12
\][/tex]
[tex]\[
f(x) = -((x - 2)^2 - 4) + 12
\][/tex]
- Simplify by distributing the \(-1\):
[tex]\[
f(x) = -(x - 2)^2 + 4 + 12
\][/tex]
[tex]\[
f(x) = -(x - 2)^2 + 16
\][/tex]
So, the vertex form of the quadratic equation is:
[tex]\[
f(x) = -(x - 2)^2 + 16
\][/tex]
Therefore, the correct choice is:
[tex]\[
f(x) = -(x - 2)^2 + 16
\][/tex]
