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Chapter 8: Problem 66

Determine the coordinates of the vertex of the graph of each function using the vertex formula. Then determine the \(x\) - and \(y\) -intercepts of the graph. Finally, plot several points and complete the graph. See Example \(9 .\) $$ f(x)=-x^{2}-2 x-1 $$

Short Answer

Expert verified
The vertex is \((-1, 0)\), the \(x\)-intercept is \((-1, 0)\), and the \(y\)-intercept is \((0, -1)\).

Step by step solution

01

Identify the Quadratic Function

The given function is a quadratic function in the form \( f(x) = ax^2 + bx + c \). For this function, \( a = -1 \), \( b = -2 \), and \( c = -1 \).
02

Use the Vertex Formula

The vertex of a quadratic function \( f(x) = ax^2 + bx + c \) is given by the formula \( x = -\frac{b}{2a} \). Substitute \( a = -1 \) and \( b = -2 \) into the formula to find the \( x \)-coordinate of the vertex:\[ x = -\frac{-2}{2(-1)} = -1 \]
03

Calculate the y-coordinate of the Vertex

Now that we have the \( x \)-coordinate of the vertex \( x = -1 \), substitute it back into the function to find the \( y \)-coordinate:\[ f(-1) = -(-1)^2 - 2(-1) - 1 = -1 + 2 - 1 = 0 \]Therefore, the vertex is \((-1, 0)\).
04

Determine the x-intercepts

The \( x \)-intercepts occur where \( f(x) = 0 \). Solve the equation \(-x^2 - 2x - 1 = 0\) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(-1)(-1)}}{2(-1)} = \frac{2 \pm \sqrt{4 - 4}}{-2} = \frac{2 \pm 0}{-2} = -1 \]The parabola touches the \( x \)-axis at one point, so the \( x \)-intercept is \((-1, 0)\).
05

Determine the y-intercept

The \( y \)-intercept occurs where \( x = 0 \). Evaluate the function at \( x = 0 \):\[ f(0) = -(0)^2 - 2(0) - 1 = -1 \]So the \( y \)-intercept is \((0, -1)\).
06

Choose and Evaluate Additional Points

Select additional \( x \)-values around the vertex and intercepts to plot the graph:- For \( x = -2 \): \[ f(-2) = -(-2)^2 - 2(-2) - 1 = -4 + 4 - 1 = -1 \]- For \( x = -3 \): \[ f(-3) = -(-3)^2 - 2(-3) - 1 = -9 + 6 - 1 = -4 \]
07

Plot the Points and Draw the Graph

Plot all the calculated points \((-1, 0)\), \((0, -1)\), \((-2, -1)\), \((-3, -4)\) on a coordinate plane. Sketch the parabolic graph smoothly through these points, noting that it opens downwards because \( a = -1 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Quadratic Functions
A quadratic function is a polynomial function of degree 2. It takes the form \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants and \( a eq 0 \). This function describes a parabola when graphed.
The sign of \( a \) is crucial. If \( a > 0 \), the parabola opens upwards, making a happy face. If \( a < 0 \), it opens downwards, like a sad face. In our exercise, \( a = -1 \), so the parabola opens downwards.
Quadratic functions are vital in various fields like physics, business, and engineering. They model phenomena like projectile motion and optimization problems.
Finding the X-Intercepts of a Quadratic Function
The \( x \)-intercepts of a quadratic function are points where the graph crosses the \( x \)-axis. At these points, the value of \( y \) is zero. To find them, set the quadratic function equal to zero and solve for \( x \).
The quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) is commonly used when factoring isn't easy. It provides solutions for \( x \)-intercepts, based on the coefficients \( a \), \( b \), and \( c \).
For our function \( -x^2 - 2x - 1 = 0 \), solving it gives \( x = -1 \). Thus, the parabola just touches the \( x \)-axis at one point: \((-1, 0)\). This is known as having a double root and means the vertex itself is the \( x \)-intercept.
Determining the Y-Intercept of a Quadratic Function
The \( y \)-intercept is where the graph crosses the \( y \)-axis. For quadratic functions, this occurs when \( x = 0 \). Plug \( x = 0 \) into the function and solve for \( f(x) \) to find the \( y \)-intercept.
In our case, substituting \( x = 0 \) into \( f(x) = -x^2 - 2x - 1 \), we get \( f(0) = -1 \). So, the \( y \)-intercept is at \((0, -1)\).
This point is handy in graphing the function because it gives a clear view of where the parabola starts on the \( y \)-axis. Knowing both \( x \)- and \( y \)-intercepts with the vertex helps in accurately sketching the graph.

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Most popular questions from this chapter

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