Answer
(a) See below.
(b) Domain: $(-\infty, \infty)$
Range:$[\frac{7}{8}, \infty)$
(c) Domain where function decreases:$(-\infty, -\frac{1}{4})$
Domain where function increases: $(-\frac{1}{4}, \infty)$
Work Step by Step
First, we convert to vertex form to find the range and increasing and decreasing domains easier:
$f(x)=2x^2+x+1$
$f(x)=2(x^2+\frac{1}{2}x)+1$
$f(x)=2(x^2+\frac{1}{2}x+1/16)+1-\frac{1}{8}$
$f(x)=2(x+\frac{1}{4})^2+\frac{7}{8}$
Recap:
The domain is a horizontal span from the function's smallest value of x to the function's largest value of x.
The range is a vertical span from the function's smallest value of f(x) to the function's largest value of f(x).
A function is increasing in the domain intervals where its slope is positive. On the other hand, a function is decreasing in the domain intervals where its slope is negative. In a quadratic function, we can find where the function starts or stops increasing by locating the vertex.