Whether you're a student or a professional, IDNLearn.com has answers for everyone. Ask any question and get a detailed, reliable answer from our community of experts.
Sagot :
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].
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. Thank you for choosing IDNLearn.com. We’re dedicated to providing clear answers, so visit us again for more solutions.
