QUESTION IMAGE
Question
the parabola and label all parts.
- $y = x^2 + 10x + 4$
axis of symmetry: ________ vertex
sketch:
- $y = -2x^2 + 8x - 5$
axis of symmetry: ________ vertex:
sketch:
- $y = x^2 - 2x + 0$
axis of symmetry: ________ vertex:
sketch:
- $y = -x^2 - 8x - 9$
axis of symmetry: ________ vertex:
sketch:
Let's solve each quadratic equation for the axis of symmetry and vertex. We'll use the formula for the axis of symmetry of a quadratic \( y = ax^2 + bx + c \), which is \( x = -\frac{b}{2a} \), and then find the corresponding \( y \)-value for the vertex.
Problem 6: \( y = x^2 + 10x + 4 \) (assuming the middle term is \( 10x \) from the handwritten note)
Step 1: Find the axis of symmetry
For \( y = x^2 + 10x + 4 \), \( a = 1 \), \( b = 10 \).
Using \( x = -\frac{b}{2a} \):
\( x = -\frac{10}{2(1)} = -5 \)
Step 2: Find the vertex
Substitute \( x = -5 \) into the equation to find \( y \):
\( y = (-5)^2 + 10(-5) + 4 = 25 - 50 + 4 = -21 \)
So the vertex is \( (-5, -21) \).
Problem 7: \( y = -2x^2 + 8x - 5 \)
Step 1: Find the axis of symmetry
For \( y = -2x^2 + 8x - 5 \), \( a = -2 \), \( b = 8 \).
Using \( x = -\frac{b}{2a} \):
\( x = -\frac{8}{2(-2)} = -\frac{8}{-4} = 2 \)
Step 2: Find the vertex
Substitute \( x = 2 \) into the equation:
\( y = -2(2)^2 + 8(2) - 5 = -8 + 16 - 5 = 3 \)
So the vertex is \( (2, 3) \).
Problem 8: \( y = x^2 - 2x + 0 \) (simplifies to \( y = x^2 - 2x \))
Step 1: Find the axis of symmetry
For \( y = x^2 - 2x \), \( a = 1 \), \( b = -2 \).
Using \( x = -\frac{b}{2a} \):
\( x = -\frac{-2}{2(1)} = \frac{2}{2} = 1 \)
Step 2: Find the vertex
Substitute \( x = 1 \) into the equation:
\( y = (1)^2 - 2(1) = 1 - 2 = -1 \)
So the vertex is \( (1, -1) \).
Problem 9: \( y = -x^2 - 8x - 9 \)
Step 1: Find the axis of symmetry
For \( y = -x^2 - 8x - 9 \), \( a = -1 \), \( b = -8 \).
Using \( x = -\frac{b}{2a} \):
\( x = -\frac{-8}{2(-1)} = -\frac{-8}{-2} = -4 \)
Step 2: Find the vertex
Substitute \( x = -4 \) into the equation:
\( y = -(-4)^2 - 8(-4) - 9 = -16 + 32 - 9 = 7 \)
So the vertex is \( (-4, 7) \).
Summary of Results:
| Equation | Axis of Symmetry | Vertex |
|---|---|---|
| \( y = -2x^2 + 8x - 5 \) | \( x = 2 \) | \( (2, 3) \) |
| \( y = x^2 - 2x \) | \( x = 1 \) | \( (1, -1) \) |
| \( y = -x^2 - 8x - 9 \) | \( x = -4 \) | \( (-4, 7) \) |
To sketch the parabolas:
- For \( y = x^2 + 10x + 4 \) and \( y = x^2 - 2x \), since \( a > 0 \), the parabolas open upward.
- For \( y = -2x^2 + 8x - 5 \) and \( y = -x^2 - 8x - 9 \), since \( a < 0 \), the parabolas open downward.
- Plot the vertex, then find a few other points (e.g., by plugging in \( x \)-values to the left and right of the axis of symmetry) to draw the parabola.
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Let's solve each quadratic equation for the axis of symmetry and vertex. We'll use the formula for the axis of symmetry of a quadratic \( y = ax^2 + bx + c \), which is \( x = -\frac{b}{2a} \), and then find the corresponding \( y \)-value for the vertex.
Problem 6: \( y = x^2 + 10x + 4 \) (assuming the middle term is \( 10x \) from the handwritten note)
Step 1: Find the axis of symmetry
For \( y = x^2 + 10x + 4 \), \( a = 1 \), \( b = 10 \).
Using \( x = -\frac{b}{2a} \):
\( x = -\frac{10}{2(1)} = -5 \)
Step 2: Find the vertex
Substitute \( x = -5 \) into the equation to find \( y \):
\( y = (-5)^2 + 10(-5) + 4 = 25 - 50 + 4 = -21 \)
So the vertex is \( (-5, -21) \).
Problem 7: \( y = -2x^2 + 8x - 5 \)
Step 1: Find the axis of symmetry
For \( y = -2x^2 + 8x - 5 \), \( a = -2 \), \( b = 8 \).
Using \( x = -\frac{b}{2a} \):
\( x = -\frac{8}{2(-2)} = -\frac{8}{-4} = 2 \)
Step 2: Find the vertex
Substitute \( x = 2 \) into the equation:
\( y = -2(2)^2 + 8(2) - 5 = -8 + 16 - 5 = 3 \)
So the vertex is \( (2, 3) \).
Problem 8: \( y = x^2 - 2x + 0 \) (simplifies to \( y = x^2 - 2x \))
Step 1: Find the axis of symmetry
For \( y = x^2 - 2x \), \( a = 1 \), \( b = -2 \).
Using \( x = -\frac{b}{2a} \):
\( x = -\frac{-2}{2(1)} = \frac{2}{2} = 1 \)
Step 2: Find the vertex
Substitute \( x = 1 \) into the equation:
\( y = (1)^2 - 2(1) = 1 - 2 = -1 \)
So the vertex is \( (1, -1) \).
Problem 9: \( y = -x^2 - 8x - 9 \)
Step 1: Find the axis of symmetry
For \( y = -x^2 - 8x - 9 \), \( a = -1 \), \( b = -8 \).
Using \( x = -\frac{b}{2a} \):
\( x = -\frac{-8}{2(-1)} = -\frac{-8}{-2} = -4 \)
Step 2: Find the vertex
Substitute \( x = -4 \) into the equation:
\( y = -(-4)^2 - 8(-4) - 9 = -16 + 32 - 9 = 7 \)
So the vertex is \( (-4, 7) \).
Summary of Results:
| Equation | Axis of Symmetry | Vertex |
|---|---|---|
| \( y = -2x^2 + 8x - 5 \) | \( x = 2 \) | \( (2, 3) \) |
| \( y = x^2 - 2x \) | \( x = 1 \) | \( (1, -1) \) |
| \( y = -x^2 - 8x - 9 \) | \( x = -4 \) | \( (-4, 7) \) |
To sketch the parabolas:
- For \( y = x^2 + 10x + 4 \) and \( y = x^2 - 2x \), since \( a > 0 \), the parabolas open upward.
- For \( y = -2x^2 + 8x - 5 \) and \( y = -x^2 - 8x - 9 \), since \( a < 0 \), the parabolas open downward.
- Plot the vertex, then find a few other points (e.g., by plugging in \( x \)-values to the left and right of the axis of symmetry) to draw the parabola.