which is the graph of the function $f(x)=\frac{1}{2}x^{2}+2x - 6$?

Explanation:

Step1: Identify the form of the function

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

Step2: Find the x - coordinate of the vertex

The formula for the x - coordinate of the vertex of a quadratic function $y = ax^{2}+bx + c$ is $x=-\frac{b}{2a}$. Substitute $a=\frac{1}{2}$ and $b = 2$ into the formula: $x=-\frac{2}{2\times\frac{1}{2}}=-\frac{2}{1}=-2$.

Step3: Find the y - coordinate of the vertex

Substitute $x = - 2$ into the function $f(x)=\frac{1}{2}x^{2}+2x - 6$. $f(-2)=\frac{1}{2}\times(-2)^{2}+2\times(-2)-6=\frac{1}{2}\times4-4 - 6=2-4 - 6=-8$. So the vertex is $(-2,-8)$.

Step4: Find the y - intercept

Set $x = 0$ in the function $f(x)$. Then $f(0)=\frac{1}{2}\times0^{2}+2\times0-6=-6$. The y - intercept is $(0,-6)$.

We can then match the graph based on these characteristics (up - opening parabola, vertex at $(-2,-8)$ and y - intercept at $(0,-6)$).

Answer:

The graph with an upward - opening parabola having vertex at $(-2,-8)$ and y - intercept at $(0,-6)$ (the first graph shown in the image if we assume the first graph has these characteristics).