which could be the graph of $f(x)=x^{2}+6x + 11$?

Answer

Explanation:

Step1: Identify the form of the quadratic function

The function $f(x)=x^{2}+6x + 11$ is a quadratic function in the form $y = ax^{2}+bx + c$, where $a = 1$, $b=6$, $c = 11$. Since $a=1>0$, the parabola opens upward.

Step2: Calculate the vertex - x - coordinate

The x - coordinate of the vertex of a quadratic function $y=ax^{2}+bx + c$ is given by $x=-\frac{b}{2a}$. Substituting $a = 1$ and $b = 6$ into the formula, we get $x=-\frac{6}{2\times1}=- 3$.

Step3: Calculate the vertex - y - coordinate

Substitute $x=-3$ into the function $f(x)=x^{2}+6x + 11$. Then $f(-3)=(-3)^{2}+6\times(-3)+11=9 - 18 + 11=2$. So the vertex is $(-3,2)$.

Step4: Analyze the y - intercept

The y - intercept is found by setting $x = 0$. So $f(0)=0^{2}+6\times0+11 = 11$.

The parabola opens upward, has a vertex at $(-3,2)$ and y - intercept at $(0,11)$.