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Question
21/41 graph y = (x - 3)^2 - 4
Step1: Identify the vertex form
The equation \( y=(x - 3)^2-4 \) is in vertex form of a parabola, \( y = a(x - h)^2 + k \), where \((h,k)\) is the vertex. Here, \( h = 3 \), \( k=-4 \), so the vertex is \((3,-4)\).
Step2: Determine the direction of opening
Since \( a = 1>0 \), the parabola opens upwards.
Step3: Find the axis of symmetry
The axis of symmetry is the vertical line \( x = h \), so \( x = 3 \).
Step4: Find the y - intercept
Set \( x = 0 \): \( y=(0 - 3)^2-4=9 - 4 = 5 \), so the y - intercept is \((0,5)\).
Step5: Find the x - intercepts
Set \( y = 0 \): \( 0=(x - 3)^2-4 \)
\((x - 3)^2=4\)
\(x - 3=\pm2\)
Case 1: \( x - 3 = 2\Rightarrow x=5\)
Case 2: \( x - 3=-2\Rightarrow x = 1\)
So the x - intercepts are \((1,0)\) and \((5,0)\).
Step6: Plot the points and draw the parabola
Plot the vertex \((3,-4)\), the y - intercept \((0,5)\), the x - intercepts \((1,0)\) and \((5,0)\), and use the axis of symmetry \( x = 3 \) to draw the parabola opening upwards.
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To graph \( y=(x - 3)^2-4 \):
- Vertex: \((3,-4)\) (from vertex form \( y=a(x - h)^2 + k \)).
- Direction: Opens upward (\( a = 1>0 \)).
- Axis of Symmetry: \( x = 3 \).
- Intercepts:
- Y - intercept: \((0,5)\) (when \( x = 0 \)).
- X - intercepts: \((1,0)\) and \((5,0)\) (when \( y = 0 \)).
Plot these points and draw a parabola opening upward with the vertex at \((3,-4)\), symmetric about \( x = 3 \).