Answer :
Let's graph the equation [tex]\( y = -x^2 + 6x - 8 \)[/tex] step by step. We'll identify significant points such as the roots and the vertex, and then plot these points to sketch the graph.
### 1. Finding the Roots
The roots of the quadratic equation are the x-values where [tex]\( y = 0 \)[/tex].
[tex]\[ -x^2 + 6x - 8 = 0 \][/tex]
We can solve this using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
Here, [tex]\( a = -1 \)[/tex], [tex]\( b = 6 \)[/tex], and [tex]\( c = -8 \)[/tex].
[tex]\[ x = \frac{-6 \pm \sqrt{6^2 - 4(-1)(-8)}}{2(-1)} \][/tex]
[tex]\[ x = \frac{-6 \pm \sqrt{36 - 32}}{-2} \][/tex]
[tex]\[ x = \frac{-6 \pm \sqrt{4}}{-2} \][/tex]
[tex]\[ x = \frac{-6 \pm 2}{-2} \][/tex]
Thus, the roots are:
[tex]\[ x = \frac{-6 + 2}{-2} = 2 \quad \text{and} \quad x = \frac{-6 - 2}{-2} = 4 \][/tex]
So, the roots are [tex]\( x = 2 \)[/tex] and [tex]\( x = 4 \)[/tex].
### 2. Finding the Vertex
The vertex of a parabola [tex]\( y = ax^2 + bx + c \)[/tex] is given by the formula [tex]\( x = -\frac{b}{2a} \)[/tex]:
[tex]\[ x = -\frac{6}{2(-1)} = 3 \][/tex]
Now, let's find the y-value of the vertex by substituting [tex]\( x = 3 \)[/tex] into the equation:
[tex]\[ y = -3^2 + 6(3) - 8 \][/tex]
[tex]\[ y = -9 + 18 - 8 \][/tex]
[tex]\[ y = 1 \][/tex]
So, the vertex is at [tex]\( (3, 1) \)[/tex].
### 3. Additional Points
Let's find two more points on the graph by selecting [tex]\( x = 1 \)[/tex] and [tex]\( x = 5 \)[/tex]:
For [tex]\( x = 1 \)[/tex]:
[tex]\[ y = -1^2 + 6(1) - 8 \][/tex]
[tex]\[ y = -1 + 6 - 8 \][/tex]
[tex]\[ y = -3 \][/tex]
So, the point is [tex]\( (1, -3) \)[/tex].
For [tex]\( x = 5 \)[/tex]:
[tex]\[ y = -5^2 + 6(5) - 8 \][/tex]
[tex]\[ y = -25 + 30 - 8 \][/tex]
[tex]\[ y = -3 \][/tex]
So, the point is [tex]\( (5, -3) \)[/tex].
### 4. Summary of Points
We have the following points to plot:
1. [tex]\( (2, 0) \)[/tex] - root
2. [tex]\( (4, 0) \)[/tex] - root
3. [tex]\( (3, 1) \)[/tex] - vertex
4. [tex]\( (1, -3) \)[/tex] - additional point
5. [tex]\( (5, -3) \)[/tex] - additional point
### 5. Plotting the Graph
Let's plot these points on a coordinate plane and sketch the parabola [tex]\( y = -x^2 + 6x - 8 \)[/tex].
1. Plot the point [tex]\( (2, 0) \)[/tex].
2. Plot the point [tex]\( (4, 0) \)[/tex].
3. Plot the vertex at [tex]\( (3, 1) \)[/tex].
4. Plot the point [tex]\( (1, -3) \)[/tex].
5. Plot the point [tex]\( (5, -3) \)[/tex].
After all the points are plotted, sketch a smooth curve passing through all the points. The parabola should open downwards because the coefficient of [tex]\( x^2 \)[/tex] is negative.
The resulting graph will show the shape of the quadratic function [tex]\( y = -x^2 + 6x - 8 \)[/tex] with the key points indicated.
### 1. Finding the Roots
The roots of the quadratic equation are the x-values where [tex]\( y = 0 \)[/tex].
[tex]\[ -x^2 + 6x - 8 = 0 \][/tex]
We can solve this using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
Here, [tex]\( a = -1 \)[/tex], [tex]\( b = 6 \)[/tex], and [tex]\( c = -8 \)[/tex].
[tex]\[ x = \frac{-6 \pm \sqrt{6^2 - 4(-1)(-8)}}{2(-1)} \][/tex]
[tex]\[ x = \frac{-6 \pm \sqrt{36 - 32}}{-2} \][/tex]
[tex]\[ x = \frac{-6 \pm \sqrt{4}}{-2} \][/tex]
[tex]\[ x = \frac{-6 \pm 2}{-2} \][/tex]
Thus, the roots are:
[tex]\[ x = \frac{-6 + 2}{-2} = 2 \quad \text{and} \quad x = \frac{-6 - 2}{-2} = 4 \][/tex]
So, the roots are [tex]\( x = 2 \)[/tex] and [tex]\( x = 4 \)[/tex].
### 2. Finding the Vertex
The vertex of a parabola [tex]\( y = ax^2 + bx + c \)[/tex] is given by the formula [tex]\( x = -\frac{b}{2a} \)[/tex]:
[tex]\[ x = -\frac{6}{2(-1)} = 3 \][/tex]
Now, let's find the y-value of the vertex by substituting [tex]\( x = 3 \)[/tex] into the equation:
[tex]\[ y = -3^2 + 6(3) - 8 \][/tex]
[tex]\[ y = -9 + 18 - 8 \][/tex]
[tex]\[ y = 1 \][/tex]
So, the vertex is at [tex]\( (3, 1) \)[/tex].
### 3. Additional Points
Let's find two more points on the graph by selecting [tex]\( x = 1 \)[/tex] and [tex]\( x = 5 \)[/tex]:
For [tex]\( x = 1 \)[/tex]:
[tex]\[ y = -1^2 + 6(1) - 8 \][/tex]
[tex]\[ y = -1 + 6 - 8 \][/tex]
[tex]\[ y = -3 \][/tex]
So, the point is [tex]\( (1, -3) \)[/tex].
For [tex]\( x = 5 \)[/tex]:
[tex]\[ y = -5^2 + 6(5) - 8 \][/tex]
[tex]\[ y = -25 + 30 - 8 \][/tex]
[tex]\[ y = -3 \][/tex]
So, the point is [tex]\( (5, -3) \)[/tex].
### 4. Summary of Points
We have the following points to plot:
1. [tex]\( (2, 0) \)[/tex] - root
2. [tex]\( (4, 0) \)[/tex] - root
3. [tex]\( (3, 1) \)[/tex] - vertex
4. [tex]\( (1, -3) \)[/tex] - additional point
5. [tex]\( (5, -3) \)[/tex] - additional point
### 5. Plotting the Graph
Let's plot these points on a coordinate plane and sketch the parabola [tex]\( y = -x^2 + 6x - 8 \)[/tex].
1. Plot the point [tex]\( (2, 0) \)[/tex].
2. Plot the point [tex]\( (4, 0) \)[/tex].
3. Plot the vertex at [tex]\( (3, 1) \)[/tex].
4. Plot the point [tex]\( (1, -3) \)[/tex].
5. Plot the point [tex]\( (5, -3) \)[/tex].
After all the points are plotted, sketch a smooth curve passing through all the points. The parabola should open downwards because the coefficient of [tex]\( x^2 \)[/tex] is negative.
The resulting graph will show the shape of the quadratic function [tex]\( y = -x^2 + 6x - 8 \)[/tex] with the key points indicated.
