Find the Function f(x)=a(x-h)^2+k

$f (x) = a {(x - h)}^{2} + k$

The function $F (x)$ can be found by evaluating the indefinite integral of the derivative $f (x)$ .

$F (x) = \int f (x) d x$

Split the single integral into multiple integrals.

$\int a {(x - h)}^{2} d x + \int k d x$

Since $a$ is constant with respect to $x$ , move $a$ out of the integral.

$a \int {(x - h)}^{2} d x + \int k d x$

Let $u = x - h$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

Let $u = x - h$ . Find $\frac{d u}{d x}$ .

Differentiate $x - h$ .

$\frac{d}{d x} [x - h]$

By the Sum Rule, the derivative of $x - h$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- h]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [- h]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$1 + \frac{d}{d x} [- h]$

Since $- h$ is constant with respect to $x$ , the derivative of $- h$ with respect to $x$ is $0$ .

$1 + 0$

Add $1$ and $0$ .

$1$

Rewrite the problem using $u$ and $d u$ .

$a \int u^{2} d u + \int k d x$

By the Power Rule, the integral of $u^{2}$ with respect to $u$ is $\frac{1}{3} u^{3}$ .

$a (\frac{1}{3} u^{3} + C) + \int k d x$

Apply the constant rule.

$a (\frac{1}{3} u^{3} + C) + k x + C$

Simplify.

Combine $\frac{1}{3}$ and $u^{3}$ .

$a (\frac{u^{3}}{3} + C) + k x + C$

Simplify.

$\frac{a u^{3}}{3} + k x + C$

Replace all occurrences of $u$ with $x - h$ .

$\frac{1}{3} a {(x - h)}^{3} + k x + C$

The function $F$ if derived from the integral of the derivative of the function. This is valid by the fundamental theorem of calculus.

$F (x) = \frac{1}{3} \cdot (a {(x - h)}^{3}) + k x + C$