Question

Hi. I need some help with this pkease. Thank you!Given the functions, complete the sections. a) Find the intercepts with the axes. b) Indicate the basic function that you will use to graph it. c) Identify the transformations that your graph will undergo, starting from its basic function. d) Draw the sketch showing the transformations, taking into account all the previous sections. Highlight f(x) with a pen or marker. e) Determine its domain and range.1. f(x) = 3 |x+1| - 3

Answer 1

ANSWERS

(a) y-intercept: y = 0

x-intercepts: x = 0 and x = -2

(b) f(x) = |x|

(c) Horizontal translation 1 unit left, vertical stretch by a factor of 3 and vertical translation 3 units down

(d) See explanation

(e) Domain: all real values

Range: y ≥ -3

Step-by-step explanation

(a) The intercepts with the axis are the values where the graph crosses each axis. The y-intercept occurs when x = 0,

`f(0)=3|0+1|-3=0`

And the x-intercept occurs when y = 0. In this case, one of the x-intercepts occurs at the same point as the y-intercept,

`0=3|x+1|-3`

Solve for x. Add 3 to both sides,

`\begin{gathered} 0+3=3|x+1|-3+3 \\ 3=3|x+1| \end{gathered}`

Divide both sides by 3,

`\begin{gathered} (3)/(3)=(3|x+1|)/(3) \\ 1=|x+1| \end{gathered}`

When we eliminate the absolute value, we have two results: one positive and one negative,

`\begin{gathered} -1=x+1\Rightarrow x=-2 \\ 1=x+1\Rightarrow x=0 \end{gathered}`

Hence, the x-intercepts are x = -2 and x = 0

(b) As we can see, this is an absolute value function. Therefore, the function we'd use to graph it is the absolute value of x,

`f(x)=|x|`

(c) The first transformation is given by adding 1 to x. This is a horizontal translation left 1 unit.

Then, we multiply the function by 3, which is a vertical stretch by a factor of 3, and, finally, we subtract 3 from the function, which is a vertical shift 3 units down.

(d) First, we have to graph the parent function and then, apply the transformations from part c.

The graph of the parent function, f(x) = |x| is,

We have to translate it to the left 1 unit, to get the graph of f(x) = |x + 1|,

Next, vertically stretch it by a factor of 3. To do this, we have to multiply the y-coordinates of the points by 3, to obtain the graph of f(x) = 3|x+1|,

And finally, translate it 3 units down,

(e) The domain is the set of all possible x-values for which the function exists. In this case, this is a linear and continuous function, so the domain is all real values.

The range is the set of the y-values the function can take. In this case, the range is the values of y from the vertex and greater, so the range is y ≥ -3