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Chapter 11: Problem 61

The graph of a quadratic function \(f\) opens downward and has no \(x\) -intercepts. In what quadrant(s) must the vertex lie? Explain your reasoning.

Short Answer

Expert verified
The vertex must lie in Quadrant I or Quadrant II.

Step by step solution

01

Understand the given information

The problem states that the quadratic function's graph opens downward and has no x-intercepts. From this information, identify key characteristics of the function and its graph.
02

Recognize the direction of the parabola

A quadratic function that opens downward has its leading coefficient (the coefficient of the squared term) negative. This implies the parabola is concave down.
03

Analyze the absence of x-intercepts

If there are no x-intercepts, the entire graph of the function lies either entirely above or entirely below the x-axis. Since the parabola opens downward, it must lie entirely above the x-axis.
04

Identify the vertex's position

For the graph to lie entirely above the x-axis (and open downward), the vertex, which is the highest point of the parabola, must lie above the x-axis. Hence, the vertex must have a positive y-coordinate.
05

Determine the possible quadrants

Since the vertex has a positive y-coordinate and must lie above the x-axis, it must be in either Quadrant I or Quadrant II. Quadrant I has both x and y coordinates positive, while Quadrant II has a negative x-coordinate and positive y-coordinate.

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Key Concepts

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

Vertex Position
In a quadratic function, the vertex is a crucial point. It's the highest or lowest point of the parabola, depending on the parabola's direction. For a quadratic function of the form \( ax^2 + bx + c \), the vertex can be found using the formula \( (-b/(2a), f(-b/(2a))) \). When a parabola opens downward, the vertex serves as the highest point. In the given problem, since the parabola opens downward and has no x-intercepts, the vertex must lie above the x-axis. Thus, the y-coordinate of the vertex is positive, ensuring the parabola does not cross the x-axis.
Parabola Direction
The direction in which a parabola opens depends on the sign of the leading coefficient (\(a\)) in the quadratic function \( ax^2 + bx + c \).
  • If \(a > 0\), the parabola opens upward, forming a U-shape.
  • If \(a < 0\), the parabola opens downward, forming an upside-down U-shape.
In this exercise, since the parabola opens downward, we know that \(a \) is negative. This causes the parabola to be concave down, which is essential in understanding why the vertex lies above the x-axis in this specific scenario.
X-Intercepts
X-intercepts are points where the graph of a function crosses the x-axis. For quadratic functions, these can be found by solving \( ax^2 + bx + c = 0 \). In the given problem, the quadratic function has no x-intercepts, meaning the parabola does not touch or cross the x-axis at any point. Since the parabola opens downward, the entire graph lies above the x-axis. This absence of x-intercepts implies that the vertex, as the highest point of the parabola, must remain above the x-axis to prevent the graph from dipping below it.
Quadrants
The coordinate plane is divided into four quadrants, which help identify the positions of different points based on their signs.
  • Quadrant I: Both x and y are positive.
  • Quadrant II: x is negative, y is positive.
  • Quadrant III: Both x and y are negative.
  • Quadrant IV: x is positive, y is negative.
Based on the given problem, the vertex of the parabola (with a positive y-coordinate and a downward direction) must lie either in Quadrant I or Quadrant II. Quadrant III and IV are ruled out since the vertex requires a positive y-coordinate, something found only in the first two quadrants.

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Most popular questions from this chapter

Replace the blanks in each equation with constants to complete the square and form a true equation. \(x^{2}+8 x+\)_____\(=(x+\)_____\()^{2}\)

Find the \(x\) -intercepts of the function given by \(g(x)=3 x^{2}-5 x-1\).

To prepare for Section \(11.2,\) review evaluating expressions and simplifying radical expressions (Sections \(1.8,10.3\) and \(10.8)\) Simplify. \([10.3],[10.8]\) $$ \sqrt{-25} $$

Solve. $$ c^{2}-8=0 $$

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