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Chapter 2: Problem 18

(a) Use the quadratic formula to find the zeros of \(f .\) (b) Find the maximum or minimum value of \(f(x)\) (c) Sketch the graph of \(f\) $$f(x)=-4 x^{2}+4 x-1$$

Short Answer

Expert verified
The function has one zero at \( x = \frac{1}{2} \) and a minimum value of 0.

Step by step solution

01

Identify coefficients for the quadratic formula

The quadratic equation is given by \( f(x) = ax^2 + bx + c \). For the function \( f(x) = -4x^2 + 4x - 1 \), the coefficients are \( a = -4 \), \( b = 4 \), and \( c = -1 \).
02

Apply the quadratic formula

The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute \( a = -4 \), \( b = 4 \), and \( c = -1 \) into the formula: \[ x = \frac{-4 \pm \sqrt{4^2 - 4(-4)(-1)}}{2(-4)} \].
03

Calculate discriminant

The discriminant is given by \( b^2 - 4ac \). Calculate \( 4^2 - 4(-4)(-1) = 16 - 16 = 0 \).
04

Solve for \( x \)

Since the discriminant is 0, there is one real solution. Substitute \( b = 4 \) and \( a = -4 \) into the formula \( x = \frac{-b}{2a} \): \[ x = \frac{-4}{2(-4)} = \frac{-4}{-8} = \frac{1}{2} \].
05

Determine the vertex of the parabola

The vertex form of a parabola \( ax^2 + bx + c \) is given by \( x = -\frac{b}{2a} \). For \( f(x) = -4x^2 + 4x - 1 \), calculate \( x = -\frac{4}{2(-4)} = \frac{1}{2} \).
06

Calculate value at vertex

Substitute \( x = \frac{1}{2} \) back into the function \( f(x) = -4x^2 + 4x - 1 \): \[ f(\frac{1}{2}) = -4(\frac{1}{2})^2 + 4(\frac{1}{2}) - 1 = -1 + 2 - 1 = 0 \]. Thus, the minimum value is 0.
07

Sketch the graph

The graph is a parabola opening downward (since \( a = -4 < 0 \)) with its vertex at \( (\frac{1}{2}, 0) \) which is also the point where it intersects the x-axis. The y-intercept is at \( (0, -1) \). Since the discriminant is zero, the vertex is also the point where the parabola is tangent to the x-axis.

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Key Concepts

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

Quadratic Formula
The quadratic formula is a powerful tool that helps solve quadratic equations of the form \(ax^2 + bx + c = 0\). It provides the zeros or roots of the equation, which are the values of \(x\) where the quadratic function is equal to zero.
  • The formula is: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
  • The part inside the square root, \(b^2 - 4ac\), is known as the discriminant.
By substituting the coefficients \(a = -4\), \(b = 4\), and \(c = -1\) into the formula:\[x = \frac{-4 \pm \sqrt{4^2 - 4(-4)(-1)}}{2(-4)}\]The calculation yields only one real solution because the discriminant is 0. This means there is one repeated root for this quadratic function. Calculating gives us \(x = \frac{1}{2}\). Therefore, the function has one real zero at \(x = \frac{1}{2}\).
Vertex of a Parabola
The vertex of a parabola is a significant point on the graph of a quadratic equation. It can represent the maximum or minimum value of the function, depending on the parabola's orientation.
  • The vertex is given as \(x = -\frac{b}{2a}\), derived from completing the square.
For the quadratic equation \(f(x) = -4x^2 + 4x - 1\), calculate \(x = -\frac{4}{2(-4)} = \frac{1}{2}\). Next, evaluate the function at \(x = \frac{1}{2}\):\[f\left(\frac{1}{2}\right) = -4\left(\frac{1}{2}\right)^2 + 4\left(\frac{1}{2}\right) - 1 = 0\]Thus, the vertex is \(\left(\frac{1}{2}, 0\right)\), and since the parabola opens downward (\(a = -4 < 0\)), this point is the maximum value of the function.
Discriminant
The discriminant is a crucial part of the quadratic formula that indicates the nature of the roots of a quadratic equation.
  • It is calculated as \(b^2 - 4ac\).
  • Its value helps determine the number and type of solutions:
    • If positive, there are two distinct real solutions.
    • If zero, there is exactly one real solution (repeated root).
    • If negative, there are no real solutions, but two complex ones.
For the equation \(f(x) = -4x^2 + 4x - 1\), the discriminant is\[4^2 - 4(-4)(-1) = 16 - 16 = 0\]Since the discriminant is zero, this confirms that there is one real, repeated root, specifically at \(x = \frac{1}{2}\). This also indicates the vertex lies on the x-axis, making it a point of tangency for the parabola.
Graphing Parabolas
Graphing parabolas involves understanding their shape and position based on the quadratic function. A parabola is a symmetric curve, and its graph can open upwards or downwards, depending on the coefficient \(a\).
  • If \(a > 0\), the parabola opens upwards with a minimum point.
  • If \(a < 0\), the parabola opens downwards with a maximum point.
For \(f(x) = -4x^2 + 4x - 1\), \(a = -4\) indicates that the parabola opens downward. Key characteristics to plot are:
  • The vertex at \(\left(\frac{1}{2}, 0\right)\), which is also the maximum point and the zero of the function.
  • The y-intercept, when \(x = 0\), gives \(f(0) = -1\), leading to the point \((0, -1)\).
Thus, the graph is a downward-opening parabola with its vertex touching the x-axis at \(x = \frac{1}{2}\), showing the convexity and giving the idea of symmetry over the line \(x = \frac{1}{2}\). This illustrates how understanding the orientation and key points assist in sketching accurate quadratic graphs.

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