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Chapter 8: Problem 9

Calculate the coordinates of the \(x\) - and \(y\) -intercepts for the following quadratics. a. \(y=3 x^{2}+2 x-1\) c. \(y=(5-2 x)(3+5 x)\) b. \(y=3(x-2)^{2}-1\) d. \(f(x)=x^{2}-5\)

Short Answer

Expert verified
a. x-intercepts: \( \frac{2}{3} \) and -1, y-intercept: -1; b. x-intercepts: 2\(\frac{1}{\sqrt{3}}\) and 2 - \(\frac{1}{\sqrt{3}}\), y-intercept: 11; c. x-intercepts: 2.5 and -0.6, y-intercept: 15; d. x-intercepts: \(\pm \sqrt{5}\), y-intercept: -5.

Step by step solution

01

Identify the formula for finding intercepts

The x-intercepts of a function are found by setting y (or f(x)) to 0 and solving for x. The y-intercept is found by substituting x = 0 into the equation and solving for y.
02

Calculate x-intercepts for part a

Set the equation to 0 and solve for x: Set 0 = 3x^2 + 2x - 1. Using the quadratic formula where a = 3, b = 2, and c = -1: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] \[x = \frac{-2 \pm \sqrt{4 + 12}}{6} \] \[x = \frac{-2 \pm 4}{6}\] This gives two solutions: x = \(\frac{2}{3}\) and x = -1.
03

Calculate y-intercept for part a

Substitute x = 0 into the equation: y = 3(0)^2 + 2(0) - 1 y = -1 So, the y-intercept is (0, -1).
04

Calculate x-intercepts for part b

Set the equation to 0 and solve for x: 0 = 3(x-2)^2 - 1 1 = 3(x-2)^2 \(\frac{1}{3}\) = (x-2)^2 taking square roots: \(\pm \frac{1}{\sqrt{3}} = x - 2\) This gives two solutions: x = 2+\(\frac{1}{\sqrt{3}}\) or x = 2-\(\frac{1}{\sqrt{3}}\).
05

Calculate y-intercept for part b

Substitute x = 0 into the equation: y = 3(0-2)^2 - 1 y = 3(4) - 1 y = 11 So, the y-intercept is (0, 11).
06

Calculate x-intercepts for part c

Set the equation to 0 and solve for x: 0 = (5 - 2x)(3 + 5x) This gives two solutions: 5 - 2x = 0 or 3 + 5x = 0 Solving these: x = \(\frac{5}{2}\) or x = -\(\frac{3}{5}\).
07

Calculate y-intercept for part c

Substitute x = 0 into the equation: y = (5 - 2(0))(3 + 5(0)) y = 5*3 y = 15 So, the y-intercept is (0, 15).
08

Calculate x-intercepts for part d

Set the equation to 0 and solve for x: 0 = x^2 - 5 This gives two solutions: x^2 = 5 taking square roots: x = \(\pm \sqrt{5}\).
09

Calculate y-intercept for part d

Substitute x = 0 into the equation: y = 0^2 - 5 y = -5 So, the y-intercept is (0, -5).

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Key Concepts

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

Understanding Quadratic Equations
Quadratic equations are a type of polynomial equation where the highest exponent of the variable (usually denoted as x) is 2. These equations take the general form of \[ y = ax^2 + bx + c \] where \(a\), \(b\), and \(c\) are constants. Quadratic equations graph as parabolas. The direction of the parabola (opening upwards or downwards) depends on the sign of \(a\). If \(a > 0\), the parabola opens upwards. If \(a < 0\), the parabola opens downwards. With quadratic equations, you are often interested in finding the intercepts, which are the points where the graph crosses the x-axis and y-axis. These intercepts are crucial for understanding the graph's behavior.
Exploring x-intercepts
The x-intercepts of a quadratic equation are the points where the graph crosses the x-axis. At these points, the value of y is zero. You can find x-intercepts by setting the quadratic equation to zero and solving for x. This involves solving the equation:\[ ax^2 + bx + c = 0 \]
One common method to solve this is by using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
This formula gives you two solutions, representing the points where the parabola crosses the x-axis. For example, for the equation \( y = 3x^2 + 2x - 1 \), substituting the values \(a = 3\), \(b = 2\), and \(c = -1\) into the formula gives you the x-intercepts \( x = \frac{2}{3} \) and \( x = -1 \). These solutions might sometimes be real numbers (the parabola crosses the x-axis) or complex numbers (the parabola does not cross the x-axis).
Discovering y-intercepts
The y-intercept of a quadratic equation is the point where the graph crosses the y-axis. At this point, the value of x is zero. To find the y-intercept, you simply substitute \(x = 0\) into the quadratic equation and solve for y.
For example, in the equation \( y = 3x^2 + 2x - 1 \), substituting \( x = 0 \) gives:
\[ y = 3(0)^2 + 2(0) - 1 \]
\[ y = -1 \] So the y-intercept is \((0, -1)\). The y-intercept is a single point since it is the only place where the graph crosses the y-axis.
Solving Quadratic Equations
Solving quadratic equations can be done in several ways depending on the form of the equation.
  • The **Quadratic Formula**, as previously mentioned, is one of the most reliable methods, suitable for almost all types of quadratic equations.
  • **Factoring** is another method, where you express the equation in a product of binomials. For example, \( y = (5 - 2x)(3 + 5x) \) already factored, gives x-intercepts directly by solving \( 5 - 2x = 0 \) and \( 3 + 5x = 0 \).
  • **Completing the Square** involves rewriting the quadratic equation to make it easier to solve. For example, solving \( y = 3(x - 2)^2 - 1 \) involves isolating the squared term and solving for x.

A thorough understanding of these methods not only helps in solving quadratic equations but also in grasping the graphical representation of solutions, intercepts, and the overall behavior of the parabolic graphs.

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Most popular questions from this chapter

(Graphing program optional.) a. On separate grids sketch the graphs of \(f(x)=\sqrt{-x+2}\) and \(g(x)=-\sqrt{x+2}\) b. Using interval notation, describe the domains of \(f(x)\) and \(g(x)\) c. Using interval notation, describe the ranges of \(f(x)\) and \(g(x)\) d. What is the simplest function \(h(x)\) from which both \(f(x)\) and \(g(x)\) could be created? e. Describe the transformations of \(h(x)\) to obtain \(f(x)\). Of \(h(x)\) to obtain \(g(x)\) f. Does the graph of \(f(x), g(x),\) or \(h(x)\) have any symmetries (across the \(x\) - or \(y\) -axis, or about the origin)?

Find the \(x\) -intercepts for each of the following functions. Will the vertex lie above, below, or on the \(x\) -axis? Find the vertex and sketch the graph, labeling the \(x\) -intercepts. a. \(y=(x+2)(x+1)\) d. \(y=\frac{1}{2}(x)(x-5)\) b. \(y=3(1-2 x)(x+3)\) e. \(q(x)=2(x-3)(x+2)\) c. \(y=-4(x+3)^{2}\) f. \(f(x)=-2(5-x)(3-2 x)\)

Using the given functions \(f, g,\) and \(h\) where $$ f(x)=x+1 \quad g(x)=e^{x} \quad h(x)=x-2 $$ a. Create the function \(k(x)=(f \circ g \circ h)(x)\). b. Describe the transformation from \(x\) to \(k(x)\).

(Graphing program optional.) Create a quadratic function in the vertex form \(y=a(x-h)^{2}+k,\) given the specified values for \(a\) and the vertex \((h, k) .\) Then rewrite the function in the standard form \(y=a x^{2}+b x+c .\) If available, use technology to check that the graphs of the two forms are the same. a. \(a=1,(h, k)=(2,-4)\) c. \(a=-2,(h, k)=(-3,1)\) b. \(a=-1,(h, k)=(4,3)\) d. \(a=\frac{1}{2},(h, k)=(-4,6)\)

The formula for the volume of a cone is \(V=\left(\frac{1}{3}\right) \pi r^{2} h .\) Assume you are holding a 6 -inch-high sugar cone for ice cream. a. Construct a function \(V(r)\) for the volume as a function of \(r\). Why don't you need the variable \(h\) in this case? Find \(V(1.5)\) and explain what have you found (using appropriate units). b. Evaluate \(V^{-1}(25) .\) Describe your results. What are the units attached to the number \(25 ?\) c. When dealing with abstract functions where \(f(x)=y\), we have sometimes used the convention of using \(x\) (rather than \(y\) ) as the input to the inverse function \(f^{-1}(x)\). Explain why it does not make sense to interchange \(V\) and \(r\) here to find the inverse function.
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