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Chapter 4: Problem 15

A quadratic function is given. (a) Express the quadratic function in standard form. (b) Find its vertex and its x- and y-intercept(s). (c) Sketch its graph. $$ f(x)=-x^{2}+6 x+4 $$

Short Answer

Expert verified
Standard form: \( -x^2 + 6x + 4 \). Vertex: \((3, 13)\). Intercepts: x at \( 3 \pm \sqrt{13} \), y at 4. Graph: Downward-opening parabola.

Step by step solution

01

Identify the standard form

The standard form of a quadratic function is given by \( f(x) = ax^2 + bx + c \). Identify the coefficients from the given function: \( a = -1 \), \( b = 6 \), and \( c = 4 \). The function is already in standard form: \( f(x) = -x^2 + 6x + 4 \).
02

Find the vertex

For a quadratic function \( f(x) = ax^2 + bx + c \), the x-coordinate of the vertex can be found using \( x = \frac{-b}{2a} \). Substituting the values, \( x = \frac{-6}{2(-1)} = 3 \). The y-coordinate is found by substituting \( x = 3 \) back into the function: \( f(3) = -(3)^2 + 6(3) + 4 = -9 + 18 + 4 = 13 \). Thus, the vertex is \((3, 13)\).
03

Find x-intercepts

The x-intercepts occur when \( f(x) = 0 \). Set the function equal to zero: \[-x^2 + 6x + 4 = 0\]. Solve this quadratic equation using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \( x = \frac{-6 \pm \sqrt{6^2 - 4(-1)(4)}}{2(-1)} \). \( x = \frac{-6 \pm \sqrt{36 + 16}}{-2} \). \( x = \frac{-6 \pm \sqrt{52}}{-2} \). \( x = \frac{-6 \pm 2\sqrt{13}}{-2} \). \( x = 3 \pm \sqrt{13} \). The x-intercepts are \( 3+\sqrt{13} \) and \( 3-\sqrt{13} \).
04

Find y-intercept

The y-intercept is found by evaluating the function at \( x = 0 \): \( f(0) = -0^2 + 6(0) + 4 = 4 \). Thus, the y-intercept is \( 4 \).
05

Sketch the graph

The graph of the quadratic function is a parabola opening downward (since \( a = -1 \)). Plot the vertex at \((3, 13)\), the y-intercept at \((0, 4)\), and the x-intercepts at approximately \((3+\sqrt{13}, 0)\) and \((3-\sqrt{13}, 0)\) on the coordinate plane. Draw a smooth curve passing through these points to obtain the parabola.

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Key Concepts

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

Standard Form of a Quadratic Function
The standard form of a quadratic function is an essential aspect to understand when dealing with quadratics. It is expressed as \( f(x) = ax^2 + bx + c \), where:
  • \( a \), \( b \), and \( c \) are constants,
  • \( a \) is not equal to zero, as it ensures the equation is quadratic.
For the function from our example, \( f(x) = -x^2 + 6x + 4 \), you can easily identify the coefficients:
  • \( a = -1 \)
  • \( b = 6 \)
  • \( c = 4 \)
Recognizing these values helps in maneuvering through other properties and characteristics of the quadratic function. The negative sign in \( a \) indicates the parabola opens downwards.
Vertex of a Parabola
The vertex of a parabola is the highest or lowest point on the graph, depending on whether the parabola opens downwards or upwards. To find the vertex in a quadratic equation in standard form, where \( f(x) = ax^2 + bx + c \), we use the vertex formula for the x-coordinate:\[ x = \frac{-b}{2a} \]For our example, substituting \( b = 6 \) and \( a = -1 \) gives:\[ x = \frac{-6}{2(-1)} = 3 \]The y-coordinate can be found by evaluating the function at this x-value:\[ f(3) = -(3)^2 + 6(3) + 4 = 13 \]Thus, the vertex of the parabola is at \((3, 13)\). It is a crucial point because it represents the parabola's maximum value, given the negative \( a \).
X- and Y-Intercepts
X- and y-intercepts are key points where the graph of a quadratic intersects the axes.

Finding the x-intercepts:

The x-intercepts occur where the function equals zero:\[ -x^2 + 6x + 4 = 0 \]Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we find the x-intercepts:\[ x = \frac{-6 \pm \sqrt{36 + 16}}{-2} = 3 \pm \sqrt{13} \]So, the x-intercepts are approximately \( 3 + \sqrt{13} \) and \( 3 - \sqrt{13} \).

Finding the y-intercept:

The y-intercept is found by evaluating the function at \( x = 0 \):\[ f(0) = 4 \]Thus, the graph intersects the y-axis at point \((0, 4)\). These intercepts help in constructing and understanding the parabola's graph.
Graphing Quadratics
Graphing a quadratic function involves plotting key points and sketching a parabola. Start by noting the vertex and intercepts.
  • The vertex is (3, 13).
  • The y-intercept is (0, 4).
  • The x-intercepts are approximately (3 + \( \sqrt{13} \, \), 0) and (3 - \( \sqrt{13} \, \), 0).
Given the negative coefficient \( a \), \( \{-1\} \, \), the parabola opens downwards.
To graph:
  • Start by plotting the vertex at (3, 13).
  • Mark the y-intercept at \((0, 4)\).
  • Estimate and plot the x-intercepts.
  • Draw a symmetrical, smooth curve from the vertex passing through these points.
This process visually represents the function helping to consolidate understanding of its behavior and key characteristics.

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Most popular questions from this chapter

Find the intercepts and asymptotes, and then sketch a graph of the rational function and state the domain and range. Use a graphing device to confirm your answer. $$ s(x)=\frac{x^{2}-2 x+1}{x^{3}-3 x^{2}} $$

Find all rational zeros of the polynomial, and write the polynomial in factored form. $$ P(x)=3 x^{5}-14 x^{4}-14 x^{3}+36 x^{2}+43 x+10 $$

Find all the real zeros of the polynomial. Use the quadratic formula if necessary, as in Example \(3(\mathrm{a}) .\) $$ P(x)=x^{3}-5 x^{2}+2 x+12 $$

Graph the rational function \(f,\) and determine all vertical asymptotes from your graph. Then graph \(f\) and \(g\) in a sufficiently large viewing rectangle to show that they have the same end behavior. $$ f(x)=\frac{x^{3}-2 x^{2}+16}{x-2}, g(x)=x^{2} $$

A Rational Function with No Asymptote Explain how you can tell (without graphing it) that the function $$ r(x)=\frac{x^{6}+10}{x^{4}+8 x^{2}+15} $$ has no \(x\) -intercept and no horizontal, vertical, or slant asymptote. What is its end behavior?'
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