Answer :
Certainly! Let's go through the solution step-by-step.
### Part A: Completely Factor [tex]\( f(x) \)[/tex]
The given function is:
[tex]\[ f(x) = 2x^2 - x - 10 \][/tex]
To factor [tex]\( f(x) \)[/tex] completely, we look for two numbers that multiply to [tex]\(2 \cdot (-10) = -20\)[/tex] and add to [tex]\(-1\)[/tex]. These numbers are [tex]\(-5\)[/tex] and [tex]\(4\)[/tex].
Rewriting the middle term using these two numbers:
[tex]\[ f(x) = 2x^2 - 5x + 4x - 10 \][/tex]
Grouping the terms:
[tex]\[ f(x) = (2x^2 - 5x) + (4x - 10) \][/tex]
Factoring out the common terms in each group:
[tex]\[ f(x) = x(2x - 5) + 2(2x - 5) \][/tex]
Factor out the common binomial factor [tex]\((2x - 5)\)[/tex]:
[tex]\[ f(x) = (x + 2)(2x - 5) \][/tex]
So, the completely factored form of [tex]\( f(x) \)[/tex] is:
[tex]\[ f(x) = (x + 2)(2x - 5) \][/tex]
### Part B: Find the [tex]\( x \)[/tex]-Intercepts
To find the [tex]\( x \)[/tex]-intercepts, set [tex]\( f(x) = 0 \)[/tex] and solve for [tex]\( x \)[/tex]:
[tex]\[ (x + 2)(2x - 5) = 0 \][/tex]
Setting each factor to zero:
[tex]\[ x + 2 = 0 \implies x = -2 \][/tex]
[tex]\[ 2x - 5 = 0 \implies 2x = 5 \implies x = \frac{5}{2} \][/tex]
So, the [tex]\( x \)[/tex]-intercepts are:
[tex]\[ x = -2 \quad \text{and} \quad x = \frac{5}{2} \][/tex]
### Part C: Describe the End Behavior of the Graph of [tex]\( f(x) \)[/tex]
The function [tex]\( f(x) = 2x^2 - x - 10 \)[/tex] is a quadratic function of the form [tex]\( ax^2 + bx + c \)[/tex], where [tex]\( a = 2 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -10 \)[/tex].
For a quadratic function [tex]\( ax^2 + bx + c \)[/tex] where [tex]\( a > 0 \)[/tex]:
- As [tex]\( x \to +\infty \)[/tex], [tex]\( f(x) \to +\infty \)[/tex]
- As [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) \to +\infty \)[/tex]
Since [tex]\( a = 2 \)[/tex] is positive, the end behavior of the graph is:
[tex]\[ \text{As } x \to \pm\infty, \ f(x) \to +\infty \][/tex]
### Part D: Steps to Graph [tex]\( f(x) \)[/tex]
Here are the steps to graph the function [tex]\( f(x) = 2x^2 - x - 10 \)[/tex]:
1. Identify the Zeros of [tex]\( f(x) \)[/tex]:
The [tex]\( x \)[/tex]-intercepts (zeros) found in Part B are [tex]\( x = -2 \)[/tex] and [tex]\( x = \frac{5}{2} \)[/tex].
2. Identify the Vertex of the Parabola:
The vertex form of a quadratic function can be found using [tex]\( x = -\frac{b}{2a} \)[/tex]. For [tex]\( f(x) \)[/tex], [tex]\( a = 2 \)[/tex] and [tex]\( b = -1 \)[/tex]:
[tex]\[ x = -\frac{-1}{2 \cdot 2} = \frac{1}{4} \][/tex]
Substitute [tex]\( x = \frac{1}{4} \)[/tex] back into [tex]\( f(x) \)[/tex] to find the [tex]\( y \)[/tex]-coordinate of the vertex:
[tex]\[ f\left( \frac{1}{4} \right) = 2\left(\frac{1}{4}\right)^2 - \left(\frac{1}{4}\right) - 10 = \frac{1}{8} - \frac{1}{4} - 10 = -10.125 \][/tex]
Thus, the vertex is at [tex]\(\left( \frac{1}{4}, -10.125 \right)\)[/tex].
3. Determine the Axis of Symmetry:
The axis of symmetry is the vertical line that passes through the vertex:
[tex]\[ x = \frac{1}{4} \][/tex]
4. Determine the End Behavior of the Graph:
From Part C, we know that as [tex]\( x \to \pm\infty \)[/tex], [tex]\( f(x) \to +\infty \)[/tex].
5. Plot the Vertex, [tex]\( x \)[/tex]-Intercepts, and Additional Points as Needed:
Plot the points [tex]\( (-2, 0) \)[/tex], [tex]\(\left( \frac{5}{2}, 0 \right) \)[/tex], and [tex]\(\left( \frac{1}{4}, -10.125 \right) \)[/tex]. You can also compute additional points if needed for more accuracy.
6. Draw a Smooth Curve that Passes Through These Points:
Draw the parabola by connecting these points, keeping in mind the symmetry around the axis [tex]\( x = \frac{1}{4} \)[/tex] and the end behavior.
Using the zeros, vertex, axis of symmetry, and end behavior provides a guide to sketch the graph accurately. The points ensure that the graph is plotted correctly and the end behavior indicates how the parabola behaves as [tex]\( x \)[/tex] moves towards positive or negative infinity.
This completes the detailed, step-by-step solution for the question using [tex]\( f(x) = 2x^2 - x - 10 \)[/tex].
### Part A: Completely Factor [tex]\( f(x) \)[/tex]
The given function is:
[tex]\[ f(x) = 2x^2 - x - 10 \][/tex]
To factor [tex]\( f(x) \)[/tex] completely, we look for two numbers that multiply to [tex]\(2 \cdot (-10) = -20\)[/tex] and add to [tex]\(-1\)[/tex]. These numbers are [tex]\(-5\)[/tex] and [tex]\(4\)[/tex].
Rewriting the middle term using these two numbers:
[tex]\[ f(x) = 2x^2 - 5x + 4x - 10 \][/tex]
Grouping the terms:
[tex]\[ f(x) = (2x^2 - 5x) + (4x - 10) \][/tex]
Factoring out the common terms in each group:
[tex]\[ f(x) = x(2x - 5) + 2(2x - 5) \][/tex]
Factor out the common binomial factor [tex]\((2x - 5)\)[/tex]:
[tex]\[ f(x) = (x + 2)(2x - 5) \][/tex]
So, the completely factored form of [tex]\( f(x) \)[/tex] is:
[tex]\[ f(x) = (x + 2)(2x - 5) \][/tex]
### Part B: Find the [tex]\( x \)[/tex]-Intercepts
To find the [tex]\( x \)[/tex]-intercepts, set [tex]\( f(x) = 0 \)[/tex] and solve for [tex]\( x \)[/tex]:
[tex]\[ (x + 2)(2x - 5) = 0 \][/tex]
Setting each factor to zero:
[tex]\[ x + 2 = 0 \implies x = -2 \][/tex]
[tex]\[ 2x - 5 = 0 \implies 2x = 5 \implies x = \frac{5}{2} \][/tex]
So, the [tex]\( x \)[/tex]-intercepts are:
[tex]\[ x = -2 \quad \text{and} \quad x = \frac{5}{2} \][/tex]
### Part C: Describe the End Behavior of the Graph of [tex]\( f(x) \)[/tex]
The function [tex]\( f(x) = 2x^2 - x - 10 \)[/tex] is a quadratic function of the form [tex]\( ax^2 + bx + c \)[/tex], where [tex]\( a = 2 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -10 \)[/tex].
For a quadratic function [tex]\( ax^2 + bx + c \)[/tex] where [tex]\( a > 0 \)[/tex]:
- As [tex]\( x \to +\infty \)[/tex], [tex]\( f(x) \to +\infty \)[/tex]
- As [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) \to +\infty \)[/tex]
Since [tex]\( a = 2 \)[/tex] is positive, the end behavior of the graph is:
[tex]\[ \text{As } x \to \pm\infty, \ f(x) \to +\infty \][/tex]
### Part D: Steps to Graph [tex]\( f(x) \)[/tex]
Here are the steps to graph the function [tex]\( f(x) = 2x^2 - x - 10 \)[/tex]:
1. Identify the Zeros of [tex]\( f(x) \)[/tex]:
The [tex]\( x \)[/tex]-intercepts (zeros) found in Part B are [tex]\( x = -2 \)[/tex] and [tex]\( x = \frac{5}{2} \)[/tex].
2. Identify the Vertex of the Parabola:
The vertex form of a quadratic function can be found using [tex]\( x = -\frac{b}{2a} \)[/tex]. For [tex]\( f(x) \)[/tex], [tex]\( a = 2 \)[/tex] and [tex]\( b = -1 \)[/tex]:
[tex]\[ x = -\frac{-1}{2 \cdot 2} = \frac{1}{4} \][/tex]
Substitute [tex]\( x = \frac{1}{4} \)[/tex] back into [tex]\( f(x) \)[/tex] to find the [tex]\( y \)[/tex]-coordinate of the vertex:
[tex]\[ f\left( \frac{1}{4} \right) = 2\left(\frac{1}{4}\right)^2 - \left(\frac{1}{4}\right) - 10 = \frac{1}{8} - \frac{1}{4} - 10 = -10.125 \][/tex]
Thus, the vertex is at [tex]\(\left( \frac{1}{4}, -10.125 \right)\)[/tex].
3. Determine the Axis of Symmetry:
The axis of symmetry is the vertical line that passes through the vertex:
[tex]\[ x = \frac{1}{4} \][/tex]
4. Determine the End Behavior of the Graph:
From Part C, we know that as [tex]\( x \to \pm\infty \)[/tex], [tex]\( f(x) \to +\infty \)[/tex].
5. Plot the Vertex, [tex]\( x \)[/tex]-Intercepts, and Additional Points as Needed:
Plot the points [tex]\( (-2, 0) \)[/tex], [tex]\(\left( \frac{5}{2}, 0 \right) \)[/tex], and [tex]\(\left( \frac{1}{4}, -10.125 \right) \)[/tex]. You can also compute additional points if needed for more accuracy.
6. Draw a Smooth Curve that Passes Through These Points:
Draw the parabola by connecting these points, keeping in mind the symmetry around the axis [tex]\( x = \frac{1}{4} \)[/tex] and the end behavior.
Using the zeros, vertex, axis of symmetry, and end behavior provides a guide to sketch the graph accurately. The points ensure that the graph is plotted correctly and the end behavior indicates how the parabola behaves as [tex]\( x \)[/tex] moves towards positive or negative infinity.
This completes the detailed, step-by-step solution for the question using [tex]\( f(x) = 2x^2 - x - 10 \)[/tex].
