IDNLearn.com is designed to help you find accurate answers with ease. Get accurate and comprehensive answers from our network of experienced professionals.
Sagot :
To graph the function \( f(x) = x^2 - 4x - 5 \), let's proceed step by step:
### Step 1: Identify \( a \) and \( b \)
The quadratic function in standard form is given by:
[tex]\[ f(x) = ax^2 + bx + c \][/tex]
For the given function \( f(x) = x^2 - 4x - 5 \):
- \( a = 1 \)
- \( b = -4 \)
So, we have:
[tex]\[ a = 1 \][/tex]
[tex]\[ b = -4 \][/tex]
### Step 2: Create a Table of Values
To graph the function, we can create a table of values by choosing several \( x \) values and calculating their corresponding \( y \) values (i.e., \( f(x) \)).
Let's choose \( x \) values: \( -2, -1, 0, 1, 2, 3, 4 \).
[tex]\[ \begin{array}{|c|c|} \hline x & y = f(x) \\ \hline -2 & (-2)^2 - 4(-2) - 5 = 4 + 8 - 5 = 7 \\ -1 & (-1)^2 - 4(-1) - 5 = 1 + 4 - 5 = 0 \\ 0 & 0^2 - 4(0) - 5 = -5 \\ 1 & 1^2 - 4(1) - 5 = 1 - 4 - 5 = -8 \\ 2 & 2^2 - 4(2) - 5 = 4 - 8 - 5 = -9 \\ 3 & 3^2 - 4(3) - 5 = 9 - 12 - 5 = -8 \\ 4 & 4^2 - 4(4) - 5 = 16 - 16 - 5 = -5 \\ \hline \end{array} \][/tex]
This table gives us pairs of \( (x, y) \) coordinates that we can plot on a graph.
### Step 3: Plot the Points and Draw the Curve
Using the table of values:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -2 & 7 \\ -1 & 0 \\ 0 & -5 \\ 1 & -8 \\ 2 & -9 \\ 3 & -8 \\ 4 & -5 \\ \hline \end{array} \][/tex]
Plot these points on the coordinate plane and draw a smooth curve passing through them. The resulting graph should be a parabola that opens upwards, given that the coefficient of \( x^2 \) (which is \( a = 1 \)) is positive.
You can also note the vertex of the parabola. The vertex form of the quadratic equation can help, but since we're doing a graphical approach, it's mainly about ensuring the shape follows the basic curve derived from the points above.
To double-check the vertex mathematically, you can use the formula for the x-coordinate of the vertex:
[tex]\[ x = -\frac{b}{2a} \][/tex]
[tex]\[ x = -\frac{-4}{2 \times 1} \][/tex]
[tex]\[ x = 2 \][/tex]
Plug \( x = 2 \) back into the function to find the vertex's y-coordinate:
[tex]\[ y = 2^2 - 4 \cdot 2 - 5 \][/tex]
[tex]\[ y = 4 - 8 - 5 \][/tex]
[tex]\[ y = -9 \][/tex]
So, the vertex of the parabola is at \( (2, -9) \), confirming our table point.
### Conclusion:
By plotting the points from the table and drawing the parabolic curve upwards, you will have the graph of the function: [tex]\( f(x) = x^2 - 4x - 5 \)[/tex].
### Step 1: Identify \( a \) and \( b \)
The quadratic function in standard form is given by:
[tex]\[ f(x) = ax^2 + bx + c \][/tex]
For the given function \( f(x) = x^2 - 4x - 5 \):
- \( a = 1 \)
- \( b = -4 \)
So, we have:
[tex]\[ a = 1 \][/tex]
[tex]\[ b = -4 \][/tex]
### Step 2: Create a Table of Values
To graph the function, we can create a table of values by choosing several \( x \) values and calculating their corresponding \( y \) values (i.e., \( f(x) \)).
Let's choose \( x \) values: \( -2, -1, 0, 1, 2, 3, 4 \).
[tex]\[ \begin{array}{|c|c|} \hline x & y = f(x) \\ \hline -2 & (-2)^2 - 4(-2) - 5 = 4 + 8 - 5 = 7 \\ -1 & (-1)^2 - 4(-1) - 5 = 1 + 4 - 5 = 0 \\ 0 & 0^2 - 4(0) - 5 = -5 \\ 1 & 1^2 - 4(1) - 5 = 1 - 4 - 5 = -8 \\ 2 & 2^2 - 4(2) - 5 = 4 - 8 - 5 = -9 \\ 3 & 3^2 - 4(3) - 5 = 9 - 12 - 5 = -8 \\ 4 & 4^2 - 4(4) - 5 = 16 - 16 - 5 = -5 \\ \hline \end{array} \][/tex]
This table gives us pairs of \( (x, y) \) coordinates that we can plot on a graph.
### Step 3: Plot the Points and Draw the Curve
Using the table of values:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -2 & 7 \\ -1 & 0 \\ 0 & -5 \\ 1 & -8 \\ 2 & -9 \\ 3 & -8 \\ 4 & -5 \\ \hline \end{array} \][/tex]
Plot these points on the coordinate plane and draw a smooth curve passing through them. The resulting graph should be a parabola that opens upwards, given that the coefficient of \( x^2 \) (which is \( a = 1 \)) is positive.
You can also note the vertex of the parabola. The vertex form of the quadratic equation can help, but since we're doing a graphical approach, it's mainly about ensuring the shape follows the basic curve derived from the points above.
To double-check the vertex mathematically, you can use the formula for the x-coordinate of the vertex:
[tex]\[ x = -\frac{b}{2a} \][/tex]
[tex]\[ x = -\frac{-4}{2 \times 1} \][/tex]
[tex]\[ x = 2 \][/tex]
Plug \( x = 2 \) back into the function to find the vertex's y-coordinate:
[tex]\[ y = 2^2 - 4 \cdot 2 - 5 \][/tex]
[tex]\[ y = 4 - 8 - 5 \][/tex]
[tex]\[ y = -9 \][/tex]
So, the vertex of the parabola is at \( (2, -9) \), confirming our table point.
### Conclusion:
By plotting the points from the table and drawing the parabolic curve upwards, you will have the graph of the function: [tex]\( f(x) = x^2 - 4x - 5 \)[/tex].
Thank you for using this platform to share and learn. Keep asking and answering. We appreciate every contribution you make. Thank you for visiting IDNLearn.com. We’re here to provide dependable answers, so visit us again soon.