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Sagot :
Sure, let's go through the problem step by step.
### (i) Finding the intercepts
#### y-intercept:
The y-intercept occurs when [tex]\( x = 0 \)[/tex]. Substitute [tex]\( x = 0 \)[/tex] into the equation [tex]\( y = -(2x + 1)^2 + 5 \)[/tex]:
[tex]\[ y = -(2(0) + 1)^2 + 5 = -(1)^2 + 5 = -1 + 5 = 4 \][/tex]
So, the y-intercept is [tex]\((0, 4)\)[/tex].
#### x-intercepts:
The x-intercepts occur when [tex]\( y = 0 \)[/tex]. Set [tex]\( y \)[/tex] to 0 in the equation and solve for [tex]\( x \)[/tex]:
[tex]\[ 0 = -(2x + 1)^2 + 5 \][/tex]
Rearrange and solve for [tex]\( (2x + 1)^2 \)[/tex]:
[tex]\[ (2x + 1)^2 = 5 \][/tex]
Take the square root of both sides:
[tex]\[ 2x + 1 = \pm \sqrt{5} \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ 2x + 1 = \sqrt{5} \quad \text{or} \quad 2x + 1 = -\sqrt{5} \][/tex]
Subtract 1 from both sides:
[tex]\[ 2x = \sqrt{5} - 1 \quad \text{or} \quad 2x = -\sqrt{5} - 1 \][/tex]
Divide by 2:
[tex]\[ x = \frac{\sqrt{5} - 1}{2} \quad \text{or} \quad x = \frac{-\sqrt{5} - 1}{2} \][/tex]
Evaluating these, we get:
[tex]\[ x \approx 0.618 \quad \text{or} \quad x \approx -1.618 \][/tex]
So, the x-intercepts are approximately [tex]\( (0.618, 0) \)[/tex] and [tex]\( (-1.618, 0) \)[/tex].
### (ii) Maximum point of the graph (vertex)
Since the quadratic function is given in the standard form [tex]\( y = a(x-h)^2 + k \)[/tex], the vertex (h, k) of the parabola [tex]\( y = a(2x + 1)^2 + 5 \)[/tex] can be found by rewriting the function:
In this case, the given function [tex]\( y = -(2x + 1)^2 + 5 \)[/tex] can be written as [tex]\( y = -[2(x + \frac{1}{2})]^2 + 5 \)[/tex].
Here, the vertex form [tex]\( y = a(x-h)^2 + k \)[/tex] tells us that the vertex [tex]\((h, k)\)[/tex] is at:
[tex]\[ h = -\frac{1}{2} \quad \text{and} \quad k = 5 \][/tex]
Thus, the maximum point is:
[tex]\[ (-0.5, 5) \][/tex]
### (iii) Sketch the graph
To sketch the graph of [tex]\( y = -(2x + 1)^2 + 5 \)[/tex]:
1. Vertex (Maximum Point): [tex]\((-0.5, 5)\)[/tex]
2. y-intercept: [tex]\((0, 4)\)[/tex]
3. x-intercepts: [tex]\((0.618, 0)\)[/tex] and [tex]\((-1.618, 0)\)[/tex]
4. Line of Symmetry: The line of symmetry is a vertical line passing through the vertex. This is [tex]\( x = -0.5 \)[/tex].
The graph is a downward-opening parabola with the vertex at [tex]\((-0.5, 5)\)[/tex].
### (iv) Equation of the line of symmetry
The line of symmetry of the graph of a quadratic function in the form [tex]\( y = ax^2 + bx + c \)[/tex] can be found from the vertex's x-coordinate. In this case, the vertex [tex]\((-0.5, 5)\)[/tex] gives us the line of symmetry:
[tex]\[ x = -0.5 \][/tex]
Overall, the answers to each part are:
1. y-intercept: [tex]\( (0, 4) \)[/tex]
2. x-intercepts: [tex]\( (0.618, 0) \)[/tex] and [tex]\( (-1.618, 0) \)[/tex]
3. Maximum point (vertex): [tex]\( (-0.5, 5) \)[/tex]
4. Line of symmetry: [tex]\( x = -0.5 \)[/tex]
### (i) Finding the intercepts
#### y-intercept:
The y-intercept occurs when [tex]\( x = 0 \)[/tex]. Substitute [tex]\( x = 0 \)[/tex] into the equation [tex]\( y = -(2x + 1)^2 + 5 \)[/tex]:
[tex]\[ y = -(2(0) + 1)^2 + 5 = -(1)^2 + 5 = -1 + 5 = 4 \][/tex]
So, the y-intercept is [tex]\((0, 4)\)[/tex].
#### x-intercepts:
The x-intercepts occur when [tex]\( y = 0 \)[/tex]. Set [tex]\( y \)[/tex] to 0 in the equation and solve for [tex]\( x \)[/tex]:
[tex]\[ 0 = -(2x + 1)^2 + 5 \][/tex]
Rearrange and solve for [tex]\( (2x + 1)^2 \)[/tex]:
[tex]\[ (2x + 1)^2 = 5 \][/tex]
Take the square root of both sides:
[tex]\[ 2x + 1 = \pm \sqrt{5} \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ 2x + 1 = \sqrt{5} \quad \text{or} \quad 2x + 1 = -\sqrt{5} \][/tex]
Subtract 1 from both sides:
[tex]\[ 2x = \sqrt{5} - 1 \quad \text{or} \quad 2x = -\sqrt{5} - 1 \][/tex]
Divide by 2:
[tex]\[ x = \frac{\sqrt{5} - 1}{2} \quad \text{or} \quad x = \frac{-\sqrt{5} - 1}{2} \][/tex]
Evaluating these, we get:
[tex]\[ x \approx 0.618 \quad \text{or} \quad x \approx -1.618 \][/tex]
So, the x-intercepts are approximately [tex]\( (0.618, 0) \)[/tex] and [tex]\( (-1.618, 0) \)[/tex].
### (ii) Maximum point of the graph (vertex)
Since the quadratic function is given in the standard form [tex]\( y = a(x-h)^2 + k \)[/tex], the vertex (h, k) of the parabola [tex]\( y = a(2x + 1)^2 + 5 \)[/tex] can be found by rewriting the function:
In this case, the given function [tex]\( y = -(2x + 1)^2 + 5 \)[/tex] can be written as [tex]\( y = -[2(x + \frac{1}{2})]^2 + 5 \)[/tex].
Here, the vertex form [tex]\( y = a(x-h)^2 + k \)[/tex] tells us that the vertex [tex]\((h, k)\)[/tex] is at:
[tex]\[ h = -\frac{1}{2} \quad \text{and} \quad k = 5 \][/tex]
Thus, the maximum point is:
[tex]\[ (-0.5, 5) \][/tex]
### (iii) Sketch the graph
To sketch the graph of [tex]\( y = -(2x + 1)^2 + 5 \)[/tex]:
1. Vertex (Maximum Point): [tex]\((-0.5, 5)\)[/tex]
2. y-intercept: [tex]\((0, 4)\)[/tex]
3. x-intercepts: [tex]\((0.618, 0)\)[/tex] and [tex]\((-1.618, 0)\)[/tex]
4. Line of Symmetry: The line of symmetry is a vertical line passing through the vertex. This is [tex]\( x = -0.5 \)[/tex].
The graph is a downward-opening parabola with the vertex at [tex]\((-0.5, 5)\)[/tex].
### (iv) Equation of the line of symmetry
The line of symmetry of the graph of a quadratic function in the form [tex]\( y = ax^2 + bx + c \)[/tex] can be found from the vertex's x-coordinate. In this case, the vertex [tex]\((-0.5, 5)\)[/tex] gives us the line of symmetry:
[tex]\[ x = -0.5 \][/tex]
Overall, the answers to each part are:
1. y-intercept: [tex]\( (0, 4) \)[/tex]
2. x-intercepts: [tex]\( (0.618, 0) \)[/tex] and [tex]\( (-1.618, 0) \)[/tex]
3. Maximum point (vertex): [tex]\( (-0.5, 5) \)[/tex]
4. Line of symmetry: [tex]\( x = -0.5 \)[/tex]
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