Answer :
Let's break down the given problem [tex]$y = -x^2 + 6x - 5$[/tex] and find the required aspects step-by-step.
### (a) Determine whether the parabola opens upward or downward:
To determine the direction in which the parabola opens, we look at the coefficient of the [tex]\( x^2 \)[/tex] term, which is the leading coefficient. In this equation, the coefficient of [tex]\( x^2 \)[/tex] is [tex]\(-1\)[/tex], which is negative. Therefore, the parabola opens downward.
### (b) Find the vertex:
The vertex form of a quadratic equation [tex]\(y = ax^2 + bx + c\)[/tex] can be calculated using the formula for the x-coordinate of the vertex [tex]\( x = -\frac{b}{2a} \)[/tex].
For the given equation, we have:
[tex]\[ a = -1, \quad b = 6, \quad c = -5 \][/tex]
Using the vertex formula:
[tex]\[ x_{\text{vertex}} = -\frac{b}{2a} = -\frac{6}{2(-1)} = -\frac{6}{-2} = 3 \][/tex]
Now, plug the x-coordinate of the vertex back into the equation to find the y-coordinate:
[tex]\[ y = -x^2 + 6x - 5 \][/tex]
[tex]\[ y_{\text{vertex}} = -(3)^2 + 6(3) - 5 \][/tex]
[tex]\[ y_{\text{vertex}} = -9 + 18 - 5 = 4 \][/tex]
Thus, the vertex of the parabola is [tex]\( (3.0, 4.0) \)[/tex].
### (c) Find the [tex]\( x \)[/tex]-intercept(s), if possible:
The [tex]\( x \)[/tex]-intercepts are the points where the parabola crosses the [tex]\( x \)[/tex]-axis. This occurs when [tex]\( y = 0 \)[/tex].
Set the equation to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ -x^2 + 6x - 5 = 0 \][/tex]
We solve this quadratic equation using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substitute [tex]\( a = -1 \)[/tex], [tex]\( b = 6 \)[/tex], and [tex]\( c = -5 \)[/tex]:
[tex]\[ x = \frac{-6 \pm \sqrt{(6)^2 - 4(-1)(-5)}}{2(-1)} \][/tex]
[tex]\[ x = \frac{-6 \pm \sqrt{36 - 20}}{-2} \][/tex]
[tex]\[ x = \frac{-6 \pm \sqrt{16}}{-2} \][/tex]
[tex]\[ x = \frac{-6 \pm 4}{-2} \][/tex]
This results in two solutions:
[tex]\[ x_1 = \frac{-6 + 4}{-2} = 1 \][/tex]
[tex]\[ x_2 = \frac{-6 - 4}{-2} = 5 \][/tex]
Thus, the [tex]\( x \)[/tex]-intercepts are [tex]\( (1.0, 0) \)[/tex] and [tex]\( (5.0, 0) \)[/tex].
### (d) Find the [tex]\( y \)[/tex]-intercept(s):
The [tex]\( y \)[/tex]-intercept is the point where the parabola crosses the [tex]\( y \)[/tex]-axis. This occurs when [tex]\( x = 0 \)[/tex].
Substitute [tex]\( x = 0 \)[/tex] into the equation:
[tex]\[ y = -0^2 + 6(0) - 5 \][/tex]
[tex]\[ y = -5 \][/tex]
Thus, the [tex]\( y \)[/tex]-intercept is [tex]\( (0, -5) \)[/tex].
### (e) Sketch the graph:
To sketch the graph of this parabola, we summarize the key points:
- The parabola opens downward.
- The vertex is at [tex]\( (3.0, 4.0) \)[/tex].
- The [tex]\( x \)[/tex]-intercepts are at [tex]\( (1.0, 0) \)[/tex] and [tex]\( (5.0, 0) \)[/tex].
- The [tex]\( y \)[/tex]-intercept is at [tex]\( (0, -5) \)[/tex].
With these points, you can draw a downward-opening parabola that passes through these intercepts and reaches its maximum value at the vertex.
In a coordinate plane, plot the points and draw a smooth curve to represent the parabola. The curve should start from the left, pass through the [tex]\( x \)[/tex]-intercept at [tex]\( (1.0, 0) \)[/tex], reach the peak at the vertex [tex]\( (3.0, 4.0) \)[/tex], and then pass through the [tex]\( x \)[/tex]-intercept at [tex]\( (5.0, 0) \)[/tex] and the [tex]\( y \)[/tex]-intercept at [tex]\( (0, -5) \)[/tex].
### (a) Determine whether the parabola opens upward or downward:
To determine the direction in which the parabola opens, we look at the coefficient of the [tex]\( x^2 \)[/tex] term, which is the leading coefficient. In this equation, the coefficient of [tex]\( x^2 \)[/tex] is [tex]\(-1\)[/tex], which is negative. Therefore, the parabola opens downward.
### (b) Find the vertex:
The vertex form of a quadratic equation [tex]\(y = ax^2 + bx + c\)[/tex] can be calculated using the formula for the x-coordinate of the vertex [tex]\( x = -\frac{b}{2a} \)[/tex].
For the given equation, we have:
[tex]\[ a = -1, \quad b = 6, \quad c = -5 \][/tex]
Using the vertex formula:
[tex]\[ x_{\text{vertex}} = -\frac{b}{2a} = -\frac{6}{2(-1)} = -\frac{6}{-2} = 3 \][/tex]
Now, plug the x-coordinate of the vertex back into the equation to find the y-coordinate:
[tex]\[ y = -x^2 + 6x - 5 \][/tex]
[tex]\[ y_{\text{vertex}} = -(3)^2 + 6(3) - 5 \][/tex]
[tex]\[ y_{\text{vertex}} = -9 + 18 - 5 = 4 \][/tex]
Thus, the vertex of the parabola is [tex]\( (3.0, 4.0) \)[/tex].
### (c) Find the [tex]\( x \)[/tex]-intercept(s), if possible:
The [tex]\( x \)[/tex]-intercepts are the points where the parabola crosses the [tex]\( x \)[/tex]-axis. This occurs when [tex]\( y = 0 \)[/tex].
Set the equation to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ -x^2 + 6x - 5 = 0 \][/tex]
We solve this quadratic equation using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substitute [tex]\( a = -1 \)[/tex], [tex]\( b = 6 \)[/tex], and [tex]\( c = -5 \)[/tex]:
[tex]\[ x = \frac{-6 \pm \sqrt{(6)^2 - 4(-1)(-5)}}{2(-1)} \][/tex]
[tex]\[ x = \frac{-6 \pm \sqrt{36 - 20}}{-2} \][/tex]
[tex]\[ x = \frac{-6 \pm \sqrt{16}}{-2} \][/tex]
[tex]\[ x = \frac{-6 \pm 4}{-2} \][/tex]
This results in two solutions:
[tex]\[ x_1 = \frac{-6 + 4}{-2} = 1 \][/tex]
[tex]\[ x_2 = \frac{-6 - 4}{-2} = 5 \][/tex]
Thus, the [tex]\( x \)[/tex]-intercepts are [tex]\( (1.0, 0) \)[/tex] and [tex]\( (5.0, 0) \)[/tex].
### (d) Find the [tex]\( y \)[/tex]-intercept(s):
The [tex]\( y \)[/tex]-intercept is the point where the parabola crosses the [tex]\( y \)[/tex]-axis. This occurs when [tex]\( x = 0 \)[/tex].
Substitute [tex]\( x = 0 \)[/tex] into the equation:
[tex]\[ y = -0^2 + 6(0) - 5 \][/tex]
[tex]\[ y = -5 \][/tex]
Thus, the [tex]\( y \)[/tex]-intercept is [tex]\( (0, -5) \)[/tex].
### (e) Sketch the graph:
To sketch the graph of this parabola, we summarize the key points:
- The parabola opens downward.
- The vertex is at [tex]\( (3.0, 4.0) \)[/tex].
- The [tex]\( x \)[/tex]-intercepts are at [tex]\( (1.0, 0) \)[/tex] and [tex]\( (5.0, 0) \)[/tex].
- The [tex]\( y \)[/tex]-intercept is at [tex]\( (0, -5) \)[/tex].
With these points, you can draw a downward-opening parabola that passes through these intercepts and reaches its maximum value at the vertex.
In a coordinate plane, plot the points and draw a smooth curve to represent the parabola. The curve should start from the left, pass through the [tex]\( x \)[/tex]-intercept at [tex]\( (1.0, 0) \)[/tex], reach the peak at the vertex [tex]\( (3.0, 4.0) \)[/tex], and then pass through the [tex]\( x \)[/tex]-intercept at [tex]\( (5.0, 0) \)[/tex] and the [tex]\( y \)[/tex]-intercept at [tex]\( (0, -5) \)[/tex].
