Sketch y=f(x) and then find each of the following: (a) The range and domain of definition of f(x) (b) Allx and y intercepts (c) f(0),f(10),f(5) and f(7)

## Step 1: Analyze the given function<br /><br />### No function is explicitly provided in the problem. We need a specific function $f(x)$ to proceed. Let's assume, for demonstration, that $f(x) = |x-5| - 2$. This is a common type of function involving absolute value, allowing us to illustrate the steps clearly.<br /><br />## Step 2: Sketch the graph of $f(x) = |x-5| - 2$<br /><br />### The graph of $y = |x|$ is a V-shape with the vertex at the origin. $f(x) = |x-5| - 2$ shifts the graph 5 units to the right and 2 units down. The vertex is at $(5, -2)$.<br /><br />## Step 3: Determine the domain and range<br /><br />### The domain is all possible $x$ values. Since $f(x) = |x-5| - 2$ is defined for all real numbers, the domain is $(-\infty, \infty)$. The range is all possible $y$ values. The vertex of the absolute value graph is at $(5, -2)$, and the graph opens upwards. Therefore, the range is $[-2, \infty)$.<br /><br />## Step 4: Find the x and y intercepts<br /><br />### For the x-intercept, set $y=0$: $0 = |x-5| - 2 \Rightarrow |x-5| = 2$. This gives $x-5 = 2$ or $x-5 = -2$, so $x=7$ or $x=3$. The x-intercepts are $(3,0)$ and $(7,0)$.<br />### For the y-intercept, set $x=0$: $y = |0-5| - 2 = 5 - 2 = 3$. The y-intercept is $(0,3)$.<br /><br />## Step 5: Calculate $f(0)$, $f(10)$, $f(5)$, and $f(7)$<br /><br />### $f(0) = |0-5| - 2 = 5-2 = 3$<br />### $f(10) = |10-5| - 2 = 5-2 = 3$<br />### $f(5) = |5-5| - 2 = 0-2 = -2$<br />### $f(7) = |7-5| - 2 = 2-2 = 0$