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Chapter 4: Problem 20

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. $$ f(x)=2 x^{2}+x-6 $$

Short Answer

Expert verified
The standard form is \( f(x) = 2x^2 + x - 6 \), with vertex \( \left(-\frac{1}{4}, -\frac{49}{8}\right) \), x-intercepts at \( \frac{3}{2} \) and \( -2 \), and y-intercept at \( -6 \).

Step by step solution

01

Write the Quadratic in Standard Form

The function given is already in standard form, which is expressed as \( f(x) = ax^2 + bx + c \). Here, \( a = 2 \), \( b = 1 \), and \( c = -6 \).
02

Find the Vertex

The vertex \((h, k)\) of a quadratic function \( ax^2 + bx + c \) can be found using the formula \( h = -\frac{b}{2a} \). Substituting in the values, \( h = -\frac{1}{2(2)} = -\frac{1}{4} \). To find \( k \), evaluate \( f(h) \):\[ f\left(-\frac{1}{4}\right) = 2\left(-\frac{1}{4}\right)^2 + \left(-\frac{1}{4}\right) - 6 \]\[ f\left(-\frac{1}{4}\right) = 2\cdot\frac{1}{16} - \frac{1}{4} - 6 \]\[ = \frac{1}{8} - \frac{2}{8} - \frac{48}{8} = -\frac{49}{8} \].Thus, the vertex is \( \left(-\frac{1}{4}, -\frac{49}{8}\right) \).
03

Find the x-intercepts

To find the x-intercepts, solve the equation \( 2x^2 + x - 6 = 0 \) using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Substitute \( a = 2 \), \( b = 1 \), \( c = -6 \) into the formula:\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-6)}}{4} \]\[ x = \frac{-1 \pm \sqrt{1 + 48}}{4} \]\[ x = \frac{-1 \pm \sqrt{49}}{4} \]\[ x = \frac{-1 \pm 7}{4} \]\[ x_1 = \frac{6}{4} = \frac{3}{2}, \ x_2 = \frac{-8}{4} = -2 \]Thus, the x-intercepts are \( x = \frac{3}{2} \) and \( x = -2 \).
04

Find the y-intercept

The y-intercept occurs where \( x = 0 \). Substitute \( x = 0 \) into the quadratic:\[ f(0) = 2(0)^2 + 1(0) - 6 = -6 \]Thus, the y-intercept is \( y = -6 \).
05

Sketch the Graph

To sketch the graph, plot the vertex \( \left(-\frac{1}{4}, -\frac{49}{8}\right) \), x-intercepts \( \frac{3}{2} \) and \( -2 \), and the y-intercept \( -6 \). Since \( a = 2 \), the parabola opens upwards. Mark these key points and draw a symmetrical parabola crossing the x-axis at \( \frac{3}{2} \) and \( -2 \), and the y-axis at \( -6 \).

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Key Concepts

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

Standard Form
A quadratic function is often expressed in what is known as the **standard form**. This form simplifies identifying the coefficients and forming the basic structure of the function. The standard form for a quadratic equation is given by:
  • \(f(x) = ax^2 + bx + c\)
Here, \(a\), \(b\), and \(c\) are constants with \(aeq 0\), where:
  • \(a\) determines the direction and width of the parabola
  • \(b\) and \(c\) adjust its position on the graph
For our example, \(f(x) = 2x^2 + x - 6\), it is already in standard form, with \(a = 2\), \(b = 1\), and \(c = -6\). Recognizing a quadratic function in this form is the first step to analyzing its graph and finding its intercepts.
Vertex
The **vertex** of a quadratic function is the point where the parabola either reaches its maximum or minimum value. This is a crucial feature because it provides the coordinates of the turning point. The vertex \((h, k)\) can be found using:
  • \(h = -\frac{b}{2a}\)
For our example:
  • \(h = -\frac{1}{2 \times 2} = -\frac{1}{4}\)
Once we calculate \(h\), we substitute it back into the original equation to find \(k\):
  • \(f\left(-\frac{1}{4}\right) = 2\left(-\frac{1}{4}\right)^2 + \left(-\frac{1}{4}\right) - 6 = -\frac{49}{8}\)
Thus, the vertex of the parabola is \(\left(-\frac{1}{4}, -\frac{49}{8}\right)\). This point is essential for understanding the shape and position of the graph.
X-intercepts
**X-intercepts** are points where the graph of a quadratic function crosses the x-axis. To find these intercepts, solve the quadratic equation by setting \(f(x) = 0\):
  • \(2x^2 + x - 6 = 0\)
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we substitute:
  • \(a = 2\), \(b = 1\), \(c = -6\)
This results in:
  • \(x = \frac{-1 \pm \sqrt{49}}{4}\)
  • \(x = \frac{-1 \pm 7}{4}\)
The solutions are \(x = \frac{3}{2}\) and \(x = -2\), identifying the points \(\frac{3}{2}\) and \(-2\) as where the graph crosses the x-axis. These points are crucial for plotting and connecting the shape of the quadratic.
Y-intercept
The **y-intercept** is the point where the graph of the quadratic function crosses the y-axis. This occurs when \(x = 0\). To find the y-intercept, simply substitute \(x = 0\) into the function:
  • \(f(0) = 2(0)^2 + 1(0) - 6 = -6\)
This tells us the graph intersects the y-axis at \(y = -6\). Identifying this intercept helps in sketching the graph by pinpointing another vital point through which the parabola passes. The y-intercept provides a starting position from which other points in the parabola's movement can be understood.
Graph Sketching
**Graph sketching** of a quadratic function involves plotting its critical points and visually assessing its shape. For a quadratic function like \(f(x) = 2x^2 + x - 6\), you start by:
  • Identifying the vertex \((-\frac{1}{4}, -\frac{49}{8})\), which represents the parabola's lowest point, because \(a = 2 > 0\) and indicates it opens upwards.
  • Marking the x-intercepts \(x = \frac{3}{2}\) and \(x = -2\), critical for showing where the graph crosses the x-axis.
  • Marking the y-intercept at \(y = -6\), where the parabola crosses the y-axis.
Next, ensure the parabola is symmetrical around the vertex and connects through the intercepts smoothly. The vertex and intercepts guide the overall drawing, confirming how the function behaves as \(x\) moves to positive and negative infinity.

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Most popular questions from this chapter

Find all the real zeros of the polynomial. Use the quadratic formula if necessary, as in Example \(3(\mathrm{a}) .\) $$ P(x)=x^{4}+2 x^{3}-2 x^{2}-3 x+2 $$

A polynomial \(P\) is given. (a) Find all the real zeros of \(P\) (b) Sketch the graph of \(P\) . $$ P(x)=3 x^{3}+17 x^{2}+21 x-9 $$

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Find all the real zeros of the polynomial. Use the quadratic formula if necessary, as in Example \(3(\mathrm{a}) .\) $$ P(x)=x^{4}-7 x^{3}+14 x^{2}-3 x-9 $$
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