Answer :
Let's start by graphing the function [tex]\( y = -x^2 + 4x - 3 \)[/tex]. This is a quadratic function, and its graph will be a parabola.
### Step 1: Identify the Coefficients
First, let's identify the coefficients of the quadratic equation:
[tex]\[ a = -1, \quad b = 4, \quad c = -3 \][/tex]
### Step 2: Find the Vertex
The vertex form of a quadratic equation [tex]\( y = ax^2 + bx + c \)[/tex] can be useful. The x-coordinate of the vertex can be found using:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Plugging in the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ x = -\frac{4}{2(-1)} = 2 \][/tex]
To find the y-coordinate of the vertex, substitute [tex]\( x = 2 \)[/tex] back into the equation:
[tex]\[ y = -(2)^2 + 4(2) - 3 \][/tex]
[tex]\[ y = -4 + 8 - 3 = 1 \][/tex]
So, the vertex of the parabola is [tex]\( (2, 1) \)[/tex].
### Step 3: Find the Zeros
To find the zeros of the function (the points where the graph intersects the x-axis), we need to solve the equation [tex]\( -x^2 + 4x - 3 = 0 \)[/tex].
The quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] helps us find the zeros.
Let's substitute in the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ x = \frac{-4 \pm \sqrt{4^2 - 4(-1)(-3)}}{2(-1)} \][/tex]
[tex]\[ x = \frac{-4 \pm \sqrt{16 - 12}}{-2} \][/tex]
[tex]\[ x = \frac{-4 \pm \sqrt{4}}{-2} \][/tex]
[tex]\[ x = \frac{-4 \pm 2}{-2} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{-4 + 2}{-2} = 1 \][/tex]
[tex]\[ x = \frac{-4 - 2}{-2} = 3 \][/tex]
Thus, the zeros of the function are [tex]\( x = 1 \)[/tex] and [tex]\( x = 3 \)[/tex].
### Step 4: Graph the Parabola
Now, let's summarize the key points to graph:
- The zeros are at [tex]\( (1, 0) \)[/tex] and [tex]\( (3, 0) \)[/tex].
- The vertex is at [tex]\( (2, 1) \)[/tex].
When you sketch the parabola, make sure it opens downwards (since the coefficient of [tex]\( x^2 \)[/tex] is negative) and passes through the points found above.
### Answer Choices
If given multiple choice options, the correct graph will be the one that opens downwards, has its vertex at [tex]\( (2, 1) \)[/tex], and zeros at [tex]\( x = 1 \)[/tex] and [tex]\( x = 3 \)[/tex]. Given this, the correct graph matches the description and indicates the zeros correctly as [tex]\( x = 1 \)[/tex] and [tex]\( x = 3 \)[/tex].
### Step 1: Identify the Coefficients
First, let's identify the coefficients of the quadratic equation:
[tex]\[ a = -1, \quad b = 4, \quad c = -3 \][/tex]
### Step 2: Find the Vertex
The vertex form of a quadratic equation [tex]\( y = ax^2 + bx + c \)[/tex] can be useful. The x-coordinate of the vertex can be found using:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Plugging in the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ x = -\frac{4}{2(-1)} = 2 \][/tex]
To find the y-coordinate of the vertex, substitute [tex]\( x = 2 \)[/tex] back into the equation:
[tex]\[ y = -(2)^2 + 4(2) - 3 \][/tex]
[tex]\[ y = -4 + 8 - 3 = 1 \][/tex]
So, the vertex of the parabola is [tex]\( (2, 1) \)[/tex].
### Step 3: Find the Zeros
To find the zeros of the function (the points where the graph intersects the x-axis), we need to solve the equation [tex]\( -x^2 + 4x - 3 = 0 \)[/tex].
The quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] helps us find the zeros.
Let's substitute in the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ x = \frac{-4 \pm \sqrt{4^2 - 4(-1)(-3)}}{2(-1)} \][/tex]
[tex]\[ x = \frac{-4 \pm \sqrt{16 - 12}}{-2} \][/tex]
[tex]\[ x = \frac{-4 \pm \sqrt{4}}{-2} \][/tex]
[tex]\[ x = \frac{-4 \pm 2}{-2} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{-4 + 2}{-2} = 1 \][/tex]
[tex]\[ x = \frac{-4 - 2}{-2} = 3 \][/tex]
Thus, the zeros of the function are [tex]\( x = 1 \)[/tex] and [tex]\( x = 3 \)[/tex].
### Step 4: Graph the Parabola
Now, let's summarize the key points to graph:
- The zeros are at [tex]\( (1, 0) \)[/tex] and [tex]\( (3, 0) \)[/tex].
- The vertex is at [tex]\( (2, 1) \)[/tex].
When you sketch the parabola, make sure it opens downwards (since the coefficient of [tex]\( x^2 \)[/tex] is negative) and passes through the points found above.
### Answer Choices
If given multiple choice options, the correct graph will be the one that opens downwards, has its vertex at [tex]\( (2, 1) \)[/tex], and zeros at [tex]\( x = 1 \)[/tex] and [tex]\( x = 3 \)[/tex]. Given this, the correct graph matches the description and indicates the zeros correctly as [tex]\( x = 1 \)[/tex] and [tex]\( x = 3 \)[/tex].
