Answer :
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].
