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Chapter 11: Problem 37

Find the vertex of the graph of each quadratic function. Determine whether the graph opens upward or downward, find any intercepts, and sketch the graph. See Examples I through \(4 .\) (IMAGE CANNOT COPY) \(f(x)=-x^{2}+2 x-12\)

Short Answer

Expert verified
The vertex is \( (1, -11) \); the parabola opens downward with the y-intercept at \( (0, -12) \) and no x-intercepts.

Step by step solution

01

Identify the Form of the Quadratic Equation

The quadratic function given is in the form \( f(x) = ax^2 + bx + c \). Here, \( f(x) = -x^2 + 2x - 12 \), where \( a = -1 \), \( b = 2 \), and \( c = -12 \).
02

Determine the Direction of the Parabola

Since the coefficient \( a = -1 \) is negative, the graph of the quadratic function opens downwards.
03

Find the Vertex of the Quadratic Function

The vertex of a quadratic function \( ax^2 + bx + c \) can be found using the formula \( x = \frac{-b}{2a} \). Substituting \( b = 2 \) and \( a = -1 \): \( x = \frac{-2}{2 \times -1} = 1 \). To find the \( y \)-coordinate of the vertex, substitute \( x = 1 \) back into the function: \( f(1) = -(1)^2 + 2 \times 1 - 12 = -1 + 2 - 12 = -11 \). Thus, the vertex is \( (1, -11) \).
04

Find the Intercepts

To find the \( y \)-intercept, substitute \( x = 0 \) into the function: \( f(0) = -0^2 + 2\times0 - 12 = -12 \). The \( y \)-intercept is at \( (0, -12) \).To find the \( x \)-intercepts, solve \( -x^2 + 2x - 12 = 0 \). This can be done by using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here: \( x = \frac{-2 \pm \sqrt{2^2 - 4\times(-1)\times(-12)}}{2\times(-1)} = \frac{-2 \pm \sqrt{4 - 48}}{-2} \) which simplifies to \( x = \frac{-2 \pm \sqrt{-44}}{-2} \). Since there is no real solution, as the discriminant is negative, there are no real \( x \)-intercepts.
05

Sketch the Graph

Begin by plotting the vertex \( (1, -11) \). Since the graph opens downward, draw a parabola that curves downwards from the vertex. Mark the y-intercept at \( (0, -12) \). Note that there are no x-intercepts as the parabola does not cross the x-axis.

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Key Concepts

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

Parabola
A quadratic function like \( f(x) = ax^2 + bx + c \) results in a special U-shaped graph known as a parabola. The parabola can open either upward or downward depending on the sign of the leading coefficient, \( a \).
  • If \( a > 0 \), the parabola opens upward.
  • If \( a < 0 \), it opens downward.
In the given function \( f(x) = -x^2 + 2x - 12 \), the coefficient \( a \) is \( -1 \), which is negative. This means the parabola opens downward, indicating that any maximum point will be at the vertex and the arms will extend downward towards negative infinity. Recognizing how the graph opens helps predict the parabola's shape and direction when sketching it.
Vertex
The vertex of a parabola is a crucial point that represents its peak or trough, depending on its direction. For a downward-opening parabola like \( -x^2 + 2x - 12 \), the vertex is the highest point.To find the vertex, we use the vertex formula:\[x = \frac{-b}{2a}\]In our example, \( a = -1 \) and \( b = 2 \). Substitute these values into the formula to find:\[x = \frac{-2}{2(-1)} = 1\]With the \( x \)-coordinate known, find the \( y \)-coordinate by substituting back into the function:\[f(1) = (-(1)^2 + 2 \times 1 - 12) = -11\]Thus, the vertex of the parabola is at \( (1, -11) \), making it the highest point on this downward-facing curve.
Intercepts
In graph analysis, intercepts are the points where the graph intersects the axes. They provide key locations that help define the shape and position of the parabola.
  • Y-intercept: This is where the parabola crosses the y-axis. To find it, set \( x = 0 \). For our function, \( f(0) = -0^2 + 2 \times 0 - 12 = -12 \), so the y-intercept is \( (0, -12) \).
  • X-intercepts: These occur where the parabola crosses the x-axis. Solve \( ax^2 + bx + c = 0 \) for \( x \). Using the quadratic formula here:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Substitute \( a = -1 \), \( b = 2 \), and \( c = -12 \):\[x = \frac{-2 \pm \sqrt{4 - 48}}{-2}\]Since the expression under the square root (the discriminant) is negative, there are no real x-intercepts for this parabola.
Quadratic Formula
The quadratic formula is a powerful tool for finding solutions to quadratic equations. It comes in handy when you want to determine the points where the parabola touches or intersects the x-axis (the x-intercepts).This formula is expressed as:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here's what each part means:
  • \( a, b, \) and \( c \) are constants from the quadratic equation \( ax^2 + bx + c = 0 \).
  • The \( \pm \) symbol means we need to calculate twice, once with a plus and once with a minus, potentially giving two solutions.
  • The term \( b^2 - 4ac \) is called the discriminant, which indicates the nature of the roots:
    • If it鈥檚 positive, you have two distinct real roots.
    • If zero, there鈥檚 exactly one real root.
    • If negative, the roots are complex and not real.
In the case of our function, the discriminant is \( -44 \), which confirms the absence of real x-intercepts.

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Most popular questions from this chapter

Find the maximum or minimum value of each function. Approximate to two decimal places. \(f(x)=-1.9 x^{2}+5.6 x-2.7\)

Find the vertex of the graph of each quadratic function. Determine whether the graph opens upward or downward, find the \(y\) -intercept, approximate the \(x\) -intercepts to one decimal place, and sketch the graph. $$ f(x)=x^{2}-6 x+4 $$

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Sketch the graph of each function. See Section 11.5 $$ g(x)=x+2 $$
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