Q. Which graph represents the function \(h(x) = 5.5\lvert x\rvert\)?

Q: What is the general shape of the graph of h(x)=5.5\lvert x\rvert?

V-shaped absolute-value graph with vertex at the origin 0,0. It opens upward; two straight rays meet at the vertex.

Q: What are the piecewise formulas for h(x)=5.5\lvert x\rvert?

h(x)=5.5x for x\ge 0 and h(x)=-5.5x for x<0. Each branch is a linear function with slope \pm 5.5.

Q: What are the domain and range of h(x)=5.5\lvert x\rvert?

Domain: -\infty, \infty . Range: [0, \infty) because absolute value is nonnegative and scaled by 5.5 .

Q: Where are the x- and y-intercepts?

Both intercepts are at the origin: x-intercept 0 and y-intercept 0 (the vertex is at (0,0)).

Q: How does the factor 5.5 affect the graph compared to y=\lvert x\rvert?

It is a vertical stretch by factor 5.5: the graph is steeper. Points (1,1) on y=\lvert x\rvert become (1,5.5) on h(x).

Q: Is h(x)=5.5\lvert x\rvert symmetric? If so, about what?

Is h(x)=5.5\lvert x\rvert symmetric? If so, about what?

Q: How do I quickly sketch h(x)=5.5\lvert x\rvert by hand?

Plot the vertex 0,0. Plot one or two points, e.g., 1,5.5 and -1,5.5. Draw straight lines through those points to the vertex; extend lines outward. Do not use arrows.

Q: Is h(x)=5.5\lvert x\rvert a linear function?.

No. It is piecewise linear: each side is linear, but the whole function is not linear because it has a corner at the vertex.

Answer

\(h(x)=5.5\lvert x\rvert\)

The vertex is at \((0,0)\).

For \(x\ge 0\), the function becomes \(h(x)=5.5x\).

For \(x\le 0\), the function becomes \(h(x)=-5.5x\).

The graph is a V-shape centered at the origin. The right side has slope \(5.5\), and the left side has slope \(-5.5\). This graph is steeper than \(y=\lvert x\rvert\).

Detailed Explanation

Write the given function and recognize its basic type.We are given the function
\[
h(x)=5.5|x|.
\]
This is an absolute value function (a V-shaped graph) multiplied by the constant 5.5.

What to do: Identify this as a vertical scaling of the basic absolute value function f(x)=|x|.
Express h(x) as a piecewise linear function to see each branch.Recall that |x| can be written piecewise. Therefore
\[
h(x)=5.5|x|=\begin{cases}
5.5x & \text{if } x\ge 0,\\[6pt]
-5.5x & \text{if } x<0.
\end{cases}
\]

What to do: Use the piecewise form to identify the two straight-line branches of the graph: one for x≥0 and one for x<0.
Find the vertex (the corner point) of the V.Because there is no horizontal or vertical shift, the vertex is at the origin. Evaluate:
\[
h(0)=5.5|0|=0.
\]
So the vertex is at (0,0).

What to do: Mark the point (0,0) as the vertex of the graph.
Determine slopes of the two linear arms.From the piecewise form, the right-hand branch (x≥0) is the line y=5.5x, so its slope is 5.5. The left-hand branch (x<0) is the line y=-5.5x, so its slope is -5.5. Both lines pass through the origin.
What to do: Draw two rays from the origin: one going up to the right with slope 5.5 and one going up to the left with slope -5.5.
Plot a few key points to fix the shape and steepness.Compute values:
\[
h(1)=5.5,\quad h(2)=11,\quad h(-1)=5.5,\quad h(-2)=11.
\]
So the graph passes through (1,5.5), (2,11), (-1,5.5), and (-2,11).

What to do: Use these points to verify the correct steepness: the arms are much steeper than the basic y=|x| graph (which would pass through (1,1)).
State domain, range, intercepts, and transformation from the parent function.Domain: all real numbers, written as (-∞,∞). Range: y≥0, written as [0,∞). The only x- and y-intercept is (0,0). This graph is a vertical stretch of f(x)=|x| by factor 5.5 (no reflection, no shifts).
What to do: Confirm the correct graph shows domain all reals, range starting at 0, and arms steeper by factor 5.5 compared with y=|x|.
Final identification of the correct graph.The correct graph is the V-shaped absolute value graph with vertex at (0,0), arms given by the lines y=5.5x for x≥0 and y=-5.5x for x<0, passing through (1,5.5) and (-1,5.5). It is a vertical stretch of y=|x| by factor 5.5, so choose the graph whose arms are much steeper than y=|x| and that passes through the points listed.
What to do: From the multiple-choice graphs, select the one that matches these criteria: vertex at the origin, V-shape, slopes ±5.5, and points (±1,5.5).

See full solution

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Algebra FAQs

What is the general shape of the graph of \(h(x)=5.5\lvert x\rvert\)?

V-shaped absolute-value graph with vertex at the origin \(0,0\). It opens upward; two straight rays meet at the vertex.

What are the piecewise formulas for \(h(x)=5.5\lvert x\rvert\)?

\(h(x)=5.5x\) for \(x\ge 0\) and \(h(x)=-5.5x\) for \(x<0\). Each branch is a linear function with slope \(\pm 5.5\).

What are the domain and range of \(h(x)=5.5\lvert x\rvert\)?

Domain: \( -\infty, \infty \). Range: \( [0, \infty) \) because absolute value is nonnegative and scaled by \( 5.5 \).

Where are the \(x\)- and \(y\)-intercepts?

Both intercepts are at the origin: x-intercept 0 and y-intercept 0 (the vertex is at \((0,0)\)).

How does the factor 5.5 affect the graph compared to \(y=\lvert x\rvert\)?

It is a vertical stretch by factor 5.5: the graph is steeper. Points (1,1) on \(y=\lvert x\rvert\) become (1,5.5) on \(h(x)\).

Is \(h(x)=5.5\lvert x\rvert\) symmetric? If so, about what?

How do I quickly sketch \(h(x)=5.5\lvert x\rvert\) by hand?

Plot the vertex \(0,0\). Plot one or two points, e.g., \(1,5.5\) and \(-1,5.5\). Draw straight lines through those points to the vertex; extend lines outward. Do not use arrows.

Is \(h(x)=5.5\lvert x\rvert\) a linear function?.