Q. What is the Vertex of \( f(x) = |x + 8| – 3 \)?
Answer
- Find the x-coordinate.Set the inside of the absolute value to 0: x + 8 = 0, which gives x = -8.
- Find the y-coordinate.Evaluate the function at x = -8.
\[ f(-8) = |-8 + 8| – 3 = -3 \]
- State the vertex.\[ (-8, -3) \]
Detailed Explanation
Solution
- Recall the basic graph of the absolute value: the function \( g(x)=|x| \) has a V-shaped graph whose vertex is at the point \( (0,0) \). The vertex occurs where the expression inside the absolute value equals zero, because \( |u| \) is minimized when \( u=0 \).
- Look at the given function \( f(x)=|x+8|-3 \). There are two transformations applied to \( |x| \):
- The expression \( x+8 \) inside the absolute value represents a horizontal shift. Replacing \( x \) by \( x+8 \) shifts the graph of \( |x| \) left by 8 units. So the vertex moves from \( (0,0) \) to \( (-8,0) \).
- The final “\(-3\)” is a vertical shift downward by 3 units. Shifting \( (-8,0) \) down 3 units gives \( (-8,-3) \).
- Verify algebraically by finding the x-value where the inside of the absolute value is zero and then evaluating the function there:
Solve \( x+8=0 \). This gives \( x=-8 \).
Evaluate \( f(-8)=|{-8}+8|-3=|0|-3=0-3=-3 \).
- Therefore the vertex of \( f(x)=|x+8|-3 \) is the point \( (-8,-3) \).
Graph
Frequently Asked Questions
What is the vertex of f(x) = |x + 8| - 3?
The vertex is at (-8, -3). This comes from the form y = |x - h| + k with h = -8 and k = -3.
How do you find the vertex of any absolute value function?
Write it as y = a|x - h| + k. The vertex is the point (h, k). Inside shifts change h, outside shifts change k.
How do you graph f(x) = |x + 8| - 3 quickly?
Start with y = |x|, shift left 8 units, then down 3 units. Plot the vertex at (-8, -3) and draw two lines with slopes +1 (right) and -1 (left).
What is the axis of symmetry for this graph?
The axis of symmetry is the vertical line x = -8, passing through the vertex.
What is the domain and range of f(x) = |x + 8| - 3?
Domain: all real numbers. Range: y >= -3 (minimum value -3 at the vertex).
Where are the x-intercepts (zeros) of the function?
Solve |x + 8| - 3 = 0 => |x + 8| = 3. So x + 8 = ±3, giving x = -5 and x = -11. Intercepts: (-5, 0) and (-11, 0).
What is the y-intercept?
Set x = 0: f(0) = |0 + 8| - 3 = 8 - 3 = 5. Y-intercept: (0, 5).
Does the graph have maximum or minimum?
It has minimum value of -3 at the vertex (-8, -3). No maximum (it increases without bound).
How do the slopes of the two branches compare?
The right branch (x > -8) has slope +1; the left branch (x < -8) has slope -1, because = 1 here.
What happens if the absolute value has negative coefficient, e.g., y = -|x + 8| - 3?
The graph reflects vertically: it opens downward, so the vertex becomes maximum. For y = -|x + 8| - 3 the vertex is still (-8, -3), but it's maximum value of -3.
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