Chapter 10: Problem 39
In Exercises \(35-42,\) sketch the graph of the equation and label the vertex. $$y=x^{2}-4 x-1$$
Short Answer
Expert verified
Answer: The vertex of the function is (2, -5), the axis of symmetry is x = 2, the x-intercepts are \((2+\sqrt{5},0)\) and \((2-\sqrt{5},0)\), and the y-intercept is (0, -1).
Step by step solution
01
Complete the Square
To rewrite the equation in vertex form, we can complete the square for the quadratic expression. The equation is given by \(y = x^2 - 4x - 1\). Add and subtract the square of half of the coefficient of the x term inside the parenthesis and factor:
\(y = (x^2 - 4x + (-4/2)^2) - (-4/2)^2 -1\)
\(y = (x^2 - 4x + 4) - 4 -1\)
\(y = (x - 2)^2 - 5\)
Now the equation is in vertex form: \(y = a(x-h)^2+k\)
02
Identify the Vertex
The vertex of the parabola is given by the point (h,k), with h and k coming from the vertex form of the equation. In this case, h=2 and k=-5, so the vertex is (2,-5).
03
Sketch the Parabola
Now that we have the vertex, we can sketch the parabola. Knowing that the coefficient of the x squared term is positive, we can determine that the parabola opens upward.
1. Plot the vertex (2,-5) on the graph.
2. Draw the axis of symmetry, which is a vertical line that passes through the vertex (x=2).
3. The parabola is symmetric with respect to the axis of symmetry, so plot a few additional points on either side of the axis to help draw the curve.
4. Sketch the parabola connecting the vertex and the other points you plotted.
04
Find the x-intercepts and y-intercepts
To find the x-intercepts, set y to 0 and solve for x:
\(0 = (x - 2)^2 - 5\)
\(5 = (x - 2)^2\)
\(\sqrt{5} = x - 2\) or \(-\sqrt{5} = x - 2\)
\(x = 2 \pm \sqrt{5}\)
The x-intercepts are \((2+\sqrt{5},0)\) and \((2-\sqrt{5},0)\). Now, find the y-intercept by setting x to 0 and solving for y:
\(y = (0 - 2)^2 - 5\)
\(y = 4 - 5\)
\(y = -1\)
The y-intercept is \((0,-1)\).
05
Label the Vertex and Intercepts
Finally, label all the important points on the graph:
1. Label the vertex as (2,-5).
2. Label the x-intercepts as \((2+\sqrt{5},0)\) and \((2-\sqrt{5},0)\).
3. Label the y-intercept as (0,-1).
Now, the sketch of the graph is complete, with all important points labeled.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Completing the Square
Completing the square is a useful algebraic technique for rewriting quadratic equations in a way that highlights key features of their graphs. It involves manipulating the equation into a perfect square trinomial, making it easier to identify the vertex of the parabola.
Let's take the equation from our exercise: \( y = x^2 - 4x - 1 \). To complete the square:
Let's take the equation from our exercise: \( y = x^2 - 4x - 1 \). To complete the square:
- Look at the x-term, which is \(-4x\). Take half of -4, which is -2, and square it to get 4.
- Add and subtract this square (4) inside the equation to transform it into a square trinomial.
- \( y = (x^2 - 4x + 4) - 4 - 1 \)
- Notice how \( x^2 - 4x + 4 \) can be factored into \( (x - 2)^2 \).
Parabola
A parabola is a symmetrical, U-shaped curve that is defined by a quadratic equation. The standard form of a quadratic equation is \( y = ax^2 + bx + c \), and the graph of this equation results in a parabola.
The direction in which a parabola opens, either upwards or downwards, is determined by the coefficient \( a \):
The direction in which a parabola opens, either upwards or downwards, is determined by the coefficient \( a \):
- If \( a > 0 \), the parabola opens upward, like the one in our exercise.
- If \( a < 0 \), it opens downward.
Vertex Form
The vertex form of a quadratic equation is a way to express the equation that easily identifies the vertex of the parabola. It is written as \( y = a(x-h)^2 + k \), where:
- \( (h, k) \) is the vertex of the parabola.
- \( a \) indicates the direction and width of the parabola.
- \( h = 2 \) and \( k = -5 \), so the vertex is at \((2, -5)\).
- This form makes it straightforward to plot the vertex, which is a critical point in graphing.
Graphing Quadratic Equations
Graphing quadratic equations helps visualize the relationship between algebraic expressions and their graphical representations. To graph a quadratic equation, follow a series of steps:
1. **Identify the vertex:** In vertex form, this is straightforward. For \( y = (x - 2)^2 - 5 \), the vertex is \((2, -5)\).2. **Plot the vertex on a coordinate grid.**
3. **Determine the line of symmetry.** This is a vertical line that runs through the vertex (here, x = 2), dividing the parabola into mirror images.4. **Understand the direction of the opening:** Because the coefficient of \( x^2 \) is positive, our parabola opens upwards.5. **Find additional points:** To assist in plotting, calculate a couple of other points by choosing x-values near the vertex and solving for y.6. **Identify intercepts:** Calculate where the parabola crosses the x-axis (x-intercepts) and y-axis (y-intercept). For example, setting \( y \) to zero helps find x-intercepts.7. **Sketch the parabola:** Connect the points smoothly, respecting symmetry and the direction of opening.This structured method ensures a precise and visually accurate representation of quadratic equations, revealing their key characteristics at a glance.
1. **Identify the vertex:** In vertex form, this is straightforward. For \( y = (x - 2)^2 - 5 \), the vertex is \((2, -5)\).2. **Plot the vertex on a coordinate grid.**
3. **Determine the line of symmetry.** This is a vertical line that runs through the vertex (here, x = 2), dividing the parabola into mirror images.4. **Understand the direction of the opening:** Because the coefficient of \( x^2 \) is positive, our parabola opens upwards.5. **Find additional points:** To assist in plotting, calculate a couple of other points by choosing x-values near the vertex and solving for y.6. **Identify intercepts:** Calculate where the parabola crosses the x-axis (x-intercepts) and y-axis (y-intercept). For example, setting \( y \) to zero helps find x-intercepts.7. **Sketch the parabola:** Connect the points smoothly, respecting symmetry and the direction of opening.This structured method ensures a precise and visually accurate representation of quadratic equations, revealing their key characteristics at a glance.