A quadratic function $f$ is given.
\[
f(x)=x^{2}+6 x
\]
(a) Express $f$ in standard form.
\[
f(x)=
\]
(b) Find the vertex and $x$ - and $y$-interc

Step 1 ：The given function is \(f(x) = x^2 + 6x\).

Step 2 ：The standard form of a quadratic function is \(f(x) = a(x-h)^2 + k\), where \((h, k)\) is the vertex of the parabola.

Step 3 ：To convert the given function to standard form, we need to complete the square.

Step 4 ：The coefficient of \(x\) here is 6, so half of it is 3, and squaring it gives 9.

Step 5 ：We rewrite the function as \(f(x) = (x^2 + 6x + 9) - 9\). This can be written as \(f(x) = (x+3)^2 - 9\).

Step 6 ：So, the standard form of the function is \(f(x) = (x+3)^2 - 9\).

Step 7 ：The vertex of the parabola is given by the point \((h, k)\), which in this case is \((-3, -9)\).

Step 8 ：The x-intercept of the function is the value of \(x\) for which \(f(x) = 0\). We can find this by setting the function equal to zero and solving for \(x\).

Step 9 ：The y-intercept of the function is the value of \(f(x)\) when \(x = 0\). We can find this by substituting \(x = 0\) into the function.

Step 10 ：The x-intercepts of the function are \(x = -6\) and \(x = 0\). The y-intercept of the function is \(y = 0\).

Step 11 ：Final Answer: The standard form of the function is \(\boxed{f(x) = (x+3)^2 - 9}\). The vertex of the parabola is \(\boxed{(-3, -9)}\). The x-intercepts are \(\boxed{x = -6}\) and \(\boxed{x = 0}\). The y-intercept is \(\boxed{y = 0}\).