name ryan.c date section 10.1 practice math modeling - quadratic functions and their graphs find the vertex of the parabola. (ex. 2 on notes) 1. $f(x) = x^2 - 4x - 2$ 2. $f(x) = -\frac{1}{2}x^2 - 2x + 1$ 3. $f(x) = 3 - 2x^2$ vertex: __________ vertex: __________ vertex: __________ use the given graph of $f$ to evaluate the expressions. 4. $f(-2)$ and $f(0)$ 5. $f(-3)$ and $f(1)$ graphs of functions are shown here $f(-2) = $ ______ $f(-3) = $ ______ $f(0) = $ ______ $f(1) = $ ______ identify the vertex, axis of symmetry, and whether the parabola opens upward or downward. (ex. 1 on notes) 6. graph of a parabola 7. graph of a parabola vertex: ______ vertex: ______ axis of symmetry: ______ axis of symmetry: ______ opens: upward / downward opens: upward / downward

Question

Accepted Answer

### Problem 1: Find the vertex of $ f(x) = x^2 - 4x - 2 $
# Explanation:
## Step 1: Recall vertex formula for quadratic $ ax^2 + bx + c $
The x-coordinate of the vertex is $ x = -\frac{b}{2a} $. For $ f(x) = x^2 - 4x - 2 $, $ a = 1 $, $ b = -4 $. So $ x = -\frac{-4}{2(1)} = 2 $.
## Step 2: Find y-coordinate by substituting $ x = 2 $
Substitute $ x = 2 $ into $ f(x) $: $ f(2) = (2)^2 - 4(2) - 2 = 4 - 8 - 2 = -6 $.
# Answer:
Vertex: $ (2, -6) $

### Problem 2: Find the vertex of $ f(x) = -\frac{1}{2}x^2 - 2x + 1 $
# Explanation:
## Step 1: Use vertex x-coordinate formula
For $ f(x) = -\frac{1}{2}x^2 - 2x + 1 $, $ a = -\frac{1}{2} $, $ b = -2 $. $ x = -\frac{-2}{2(-\frac{1}{2})} = -\frac{-2}{-1} = -2 $.
## Step 2: Substitute $ x = -2 $ to find y
$ f(-2) = -\frac{1}{2}(-2)^2 - 2(-2) + 1 = -\frac{1}{2}(4) + 4 + 1 = -2 + 4 + 1 = 3 $.
# Answer:
Vertex: $ (-2, 3) $

### Problem 3: Find the vertex of $ f(x) = 3 - 2x^2 $ (rewritten as $ f(x) = -2x^2 + 0x + 3 $)
# Explanation:
## Step 1: Find x-coordinate of vertex
$ a = -2 $, $ b = 0 $. $ x = -\frac{0}{2(-2)} = 0 $.
## Step 2: Substitute $ x = 0 $ to find y
$ f(0) = 3 - 2(0)^2 = 3 $.
# Answer:
Vertex: $ (0, 3) $

### Problem 4: Evaluate $ f(-2) $ and $ f(0) $ from the first graph
# Explanation:
## Step 1: Find $ f(-2) $ from the graph
Locate $ x = -2 $ on the x-axis, then find the corresponding y-value on the parabola. The graph shows at $ x = -2 $, $ y = -1 $? Wait, no—wait the first graph (left) is a parabola opening upward with vertex at (0, -3)? Wait no, looking at the first graph: the parabola has vertex at (0, -3)? Wait no, the grid: x from -3 to 3, y from -3 to 3. Wait the first graph: when x = -2, the point is at y = -1? Wait no, let's re-examine. Wait the first graph: the parabola passes through (-2, -1)? Wait no, maybe I misread. Wait the first graph: the parabola has vertex at (0, -3)? Wait no, the graph is $ y = f(x) $, opening upward, vertex at (0, -3)? Wait no, the grid lines: each square is 1 unit. So at x = -2, the y-coordinate: let's see, the parabola at x = -2: moving left from x=0 (vertex), the parabola goes up. Wait the vertex is at (0, -3)? Wait no, the first graph's vertex is at (0, -3)? Wait the label is $ y = f(x) $, and the vertex is at (0, -3)? Wait no, the user's graph: first graph (left) has vertex at (0, -3)? Wait no, the original graph: "y = f(x)" with vertex at (0, -3)? Wait no, the first graph: when x = -2, the point is at y = -1? Wait no, let's check again. Wait the first graph: x=-2, the y is -1? Wait no, maybe the vertex is at (0, -3), and at x=-2, the y is -1? Wait no, let's count the grid. Each square is 1. So from vertex (0, -3), moving left 2 units (x=-2), the parabola goes up 2 units? Wait no, quadratic $ y = x^2 - 3 $ would have vertex (0, -3), and at x=-2, y = 4 - 3 = 1? Wait no, the first graph: maybe the equation is $ y = x^2 - 3 $. So at x=-2, y = (-2)^2 - 3 = 4 - 3 = 1? Wait the user's graph: the first graph has a parabola opening upward, vertex at (0, -3), and at x=-2, the point is at y=1? Wait the user's graph: "f(-2) = " and "f(0) = ". Let's re-express: the first graph: when x=0, the vertex is at (0, -3), so f(0) = -3. When x=-2, the point is at y=1? Wait no, maybe the graph is $ y = (x)^2 - 3 $, so f(0) = -3, f(-2) = (-2)^2 - 3 = 1. Wait but the user's graph: the first graph's y-axis: from -3 to 3, x from -3 to 3. So at x=-2, the y is 1, x=0 is -3. So:
## Step 1: Find $ f(-2) $
At x = -2, the y-coordinate is 1 (since the parabola $ y = x^2 - 3 $ at x=-2 is 4 - 3 = 1).
## Step 2: Find $ f(0) $
At x = 0, the vertex is at (0, -3), so f(0) = -3.
# Answer:
$ f(-2) = 1 $
$ f(0) = -3 $

### Problem 5: Evaluate $ f(-3) $ and $ f(1) $ from the second graph
# Explanation:
## Step 1: Find $ f(-3) $ from the graph
The second graph is a parabola opening downward, with vertex at (-1, 2) (approx). At x = -3, the point is at y = -3 (since the graph goes through (-3, -3)).
## Step 2: Find $ f(1) $ from the graph
At x = 1, the point is at y = -1 (since the parabola at x=1 is at y=-1).
# Answer:
$ f(-3) = -3 $
$ f(1) = -1 $

### Problem 6: Identify vertex, axis of symmetry, and direction for the third graph (left of the two at the bottom)
# Explanation:
## Step 1: Find vertex
The third graph (left bottom) is a parabola opening upward, with vertex at (1, -2) (looking at the grid: x=1, y=-2).
## Step 2: Axis of symmetry
The axis of symmetry is the vertical line through the vertex, so $ x = 1 $.
## Step 3: Direction of opening
Since the parabola opens upward (the arms go up), it opens upward.
# Answer:
Vertex: $ (1, -2) $
Axis of Symmetry: $ x = 1 $
Opens: Upward

### Problem 7: Identify vertex, axis of symmetry, and direction for the fourth graph (right bottom)
# Explanation:
## Step 1: Find vertex
The fourth graph (right bottom) is a parabola opening downward, with vertex at (-2, 3) (looking at the grid: x=-2, y=3).
## Step 2: Axis of symmetry
The axis of symmetry is the vertical line through the vertex, so $ x = -2 $.
## Step 3: Direction of opening
Since the parabola opens downward (the arms go down), it opens downward.
# Answer:
Vertex: $ (-2, 3) $
Axis of Symmetry: $ x = -2 $
Opens: Downward

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QUESTION IMAGE

Question

Explanation:

Problem 1: Find the vertex of \( f(x) = x^2 - 4x - 2 \)

Step 1: Recall vertex formula for quadratic \( ax^2 + bx + c \)

Step 2: Find y-coordinate by substituting \( x = 2 \)

Step 1: Use vertex x-coordinate formula

Step 2: Substitute \( x = -2 \) to find y

Step 1: Find x-coordinate of vertex

Step 2: Substitute \( x = 0 \) to find y

Answer:

Problem 2: Find the vertex of \( f(x) = -\frac{1}{2}x^2 - 2x + 1 \)