QUESTION IMAGE
Question
- find the vertex, axis of symmetry, and direction of opening of the parabola. use this information to sketch the graph.
(a) ( y = (x - 2)^2 + 1 )
(b) ( y = -\frac{1}{2}(x + 4)^2 + 0 )
(c) ( y = 2(x + 1)^2 - 8 )
Let's solve part (a) first: \( y = (x - 2)^2 + 1 \)
Step1: Recall vertex form of parabola
The vertex form of a parabola is \( y = a(x - h)^2 + k \), where \((h, k)\) is the vertex, the axis of symmetry is \( x = h \), and if \( a > 0 \) it opens upward, if \( a < 0 \) it opens downward.
For \( y = (x - 2)^2 + 1 \), we compare with \( y = a(x - h)^2 + k \). Here, \( a = 1 \), \( h = 2 \), \( k = 1 \).
Step2: Find vertex
From the vertex form, the vertex \((h, k)\) is \((2, 1)\).
Step3: Find axis of symmetry
The axis of symmetry is \( x = h \), so \( x = 2 \).
Step4: Find direction of opening
Since \( a = 1 > 0 \), the parabola opens upward.
Now part (b): \( y = -\frac{1}{2}(x + 4)^2 + 0 \) (assuming the \( +0 \) is just \( y = -\frac{1}{2}(x + 4)^2 \))
Step1: Recall vertex form
Using \( y = a(x - h)^2 + k \), here we can rewrite \( x + 4 \) as \( x - (-4) \), so \( h = -4 \), \( k = 0 \), \( a = -\frac{1}{2} \).
Step2: Find vertex
Vertex \((h, k)\) is \((-4, 0)\).
Step3: Find axis of symmetry
Axis of symmetry is \( x = h = -4 \).
Step4: Find direction of opening
Since \( a = -\frac{1}{2} < 0 \), the parabola opens downward.
Part (c): \( y = 2(x + 1)^2 - 8 \)
Step1: Recall vertex form
Using \( y = a(x - h)^2 + k \), rewrite \( x + 1 \) as \( x - (-1) \), so \( h = -1 \), \( k = -8 \), \( a = 2 \).
Step2: Find vertex
Vertex \((h, k)\) is \((-1, -8)\).
Step3: Find axis of symmetry
Axis of symmetry is \( x = h = -1 \).
Step4: Find direction of opening
Since \( a = 2 > 0 \), the parabola opens upward.
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s:
Part (a)
- Vertex: \((2, 1)\)
- Axis of Symmetry: \( x = 2 \)
- Direction of Opening: Upward
Part (b)
- Vertex: \((-4, 0)\)
- Axis of Symmetry: \( x = -4 \)
- Direction of Opening: Downward
Part (c)
- Vertex: \((-1, -8)\)
- Axis of Symmetry: \( x = -1 \)
- Direction of Opening: Upward