which statements about the graph of the function f(x) = 2x² - x - 6 are true? select two options. the domain…

Explanation:

Step1: Recall domain of quadratic function

For a quadratic function $f(x)=ax^{2}+bx + c$, the domain is all real - numbers. Here $a = 2$, $b=-1$, $c = - 6$, so the domain is ${x|x\in R}$, and the first option is false.

Step2: Find the vertex of the quadratic function

The $x$ - coordinate of the vertex of a quadratic function $y = ax^{2}+bx + c$ is $x=-\frac{b}{2a}$. Substitute $a = 2$ and $b=-1$ into the formula: $x=-\frac{-1}{2\times2}=\frac{1}{4}$. Then find the $y$ - coordinate by substituting $x = \frac{1}{4}$ into the function $y=f(\frac{1}{4})=2(\frac{1}{4})^{2}-\frac{1}{4}-6=2\times\frac{1}{16}-\frac{1}{4}-6=\frac{1}{8}-\frac{2}{8}-6=-6\frac{1}{8}$. The vertex is $(\frac{1}{4},-6\frac{1}{8})$.

Step3: Determine the range of the quadratic function

Since $a = 2>0$, the parabola opens upward. The minimum value of the function is the $y$ - coordinate of the vertex. So the range is ${y|y\geq - 6\frac{1}{8}}$, and the second option is false.

Step4: Find the x - intercepts

Set $y = 0$, so $2x^{2}-x - 6=0$. Factor the quadratic equation: $2x^{2}-x - 6=(2x + 3)(x - 2)=0$. Solve for $x$: $2x+3 = 0$ gives $x=-\frac{3}{2}$, and $x - 2=0$ gives $x = 2$. So the function has two $x$ - intercepts.

Step5: Determine the increasing interval

Since the parabola opens upward and the vertex is at $x=\frac{1}{4}$, the function is increasing over the interval $(\frac{1}{4},\infty)$.

Answer:

The function has two x - intercepts; The vertex of the function is $(\frac{1}{4},-6\frac{1}{8})$