To find the equation of the quadratic function in both vertex and standard form, given the vertex and a point it passes through, we follow these steps:
1. Vertex Form of the Quadratic Function:
The vertex form of a quadratic function is given by:
[tex]\[
y = a(x - h)^2 + k
\][/tex]
Here, [tex]\((h, k)\)[/tex] is the vertex of the parabola. Given the vertex [tex]\((-1, 6)\)[/tex], we can substitute [tex]\(h = -1\)[/tex] and [tex]\(k = 6\)[/tex] into the vertex form:
[tex]\[
y = a(x - (-1))^2 + 6
\][/tex]
Simplifying the expression inside the parentheses, we get:
[tex]\[
y = a(x + 1)^2 + 6
\][/tex]
2. Determine the Value of [tex]\(a\)[/tex]:
We are given a point [tex]\((2, 24)\)[/tex] that the function passes through. Substitute [tex]\(x = 2\)[/tex] and [tex]\(y = 24\)[/tex] into the equation:
[tex]\[
24 = a(2 + 1)^2 + 6
\][/tex]
Solve for [tex]\(a\)[/tex]:
[tex]\[
24 = a(3)^2 + 6
\][/tex]
[tex]\[
24 = 9a + 6
\][/tex]
Subtract 6 from both sides:
[tex]\[
18 = 9a
\][/tex]
Divide by 9:
[tex]\[
a = 2
\][/tex]
3. Write the Vertex Form:
Substitute [tex]\(a = 2\)[/tex] back into the vertex form:
[tex]\[
y = 2(x + 1)^2 + 6
\][/tex]
4. Convert to Standard Form:
To find the standard form [tex]\(y = ax^2 + bx + c\)[/tex], expand the vertex form:
[tex]\[
y = 2(x + 1)^2 + 6
\][/tex]
First, expand [tex]\((x + 1)^2\)[/tex]:
[tex]\[
(x + 1)^2 = x^2 + 2x + 1
\][/tex]
Substitute back into the equation:
[tex]\[
y = 2(x^2 + 2x + 1) + 6
\][/tex]
Distribute the 2:
[tex]\[
y = 2x^2 + 4x + 2 + 6
\][/tex]
Combine like terms:
[tex]\[
y = 2x^2 + 4x + 8
\][/tex]
Thus, the equations of the quadratic function are:
- Vertex form: [tex]\(\boxed{y = 2(x + 1)^2 + 6}\)[/tex]
- Standard form: [tex]\(\boxed{y = 2x^2 + 4x + 8}\)[/tex]
