Let's solve each question one by one.
Question 35: Find the axis of symmetry for the equation \( y = -x^2 - 4x + 1 \)
The general form of a quadratic equation is \( y = ax^2 + bx + c \). The axis of symmetry can be found using the formula:
\[ x = -\frac{b}{2a} \]
In the equation \( y = -x^2 - 4x + 1 \):
- \( a = -1 \)
- \( b = -4 \)
Plugging these values into the formula:
\[ x = -\frac{-4}{2 \cdot -1} = \frac{4}{-2} = -2 \]
So the axis of symmetry is: \[ x = -2 \]
Question 36: Find the vertex of the equation \( y = 2x^2 + 8 \)
For a quadratic in the form \( y = ax^2 + bx + c \), the vertex can also be found using the axis of symmetry. The x-coordinate of the vertex is given by:
\[ x = -\frac{b}{2a} \]
In this equation:
- \( a = 2 \)
- \( b = 0 \)
Calculating the x-coordinate of the vertex:
\[ x = -\frac{0}{2 \cdot 2} = 0 \]
To find the y-coordinate, substitute \( x = 0 \) into the equation:
\[ y = 2(0)^2 + 8 = 8 \]
Thus, the vertex is: \[ (0, 8) \]
Question 37: Identify the zeros of the equation \( y = -(x-2)^2 + 4 \)
To find the zeros, we set \( y \) to zero and solve for \( x \):
\[ 0 = -(x-2)^2 + 4 \] \[ -(x-2)^2 = -4 \] \[ (x-2)^2 = 4 \]
Taking the square root of both sides gives:
\[ x - 2 = 2 \quad \text{or} \quad x - 2 = -2 \]
Solving each equation:
- \( x - 2 = 2 \) leads to \( x = 4 \)
- \( x - 2 = -2 \) leads to \( x = 0 \)
Thus, the zeros are: \[ {0, 4} \]
Summary of Answers:
- Question 35: Axis of symmetry: \( x = -2 \)
- Question 36: Vertex: \( (0, 8) \)
- Question 37: Zeros: \( {0, 4} \)
1 answer