Section 9.5 Topics in Graphing Chapter Review
ΒΆSubsection 9.5.1 Review of Graphing
permalinkIn Section 9.1, we reviewed the fundamentals of graph-making. In particular, given an equation of the form y=expression in x, the fundamental approach to making a graph is to make a table of points to plot.
permalinkWe also looked back at the notions of βinterceptsβ on a graph. In the case of a linear equation in x and y, finding the x- and y-intercepts can be a way to create a graph.
Subsection 9.5.2 Key Features of Quadratic Graphs
permalinkIn Section 9.2, we identified the key features of a quadratic graph (which takes the shape of a parabola). The key features are the direction that it opens, the vertex, the axis of symmetry, the vertical intercept, and the horizontal intercepts (if there are any).
permalinkIf the equation for a quadratic curve is y=ax2+bx+c, then the formula h=βb2a gives the first coordinate of the vertex. So you can find the location of the vertex with that coordinate and subbing that number into the equation to find the second coordinate.
permalinkIf we know the location of a parabola's vertex and the direction that it opens, we can sketch the parabola. It helps to make a table finding a few points the the left and to the right of the vertex. The symmetry of a parabola means you only need to find points on one side to automatically get corresponding points on the other side.
Subsection 9.5.3 Graphing Quadratic Equations
permalinkIn Section 9.3, we practiced finding the exact locations of the vertical and horizontal intercepts for a quadratic equation curve. The vertical intercept can be found by lettting x=0. The result is a number on the y-axis.
permalinkThe horizontal intercepts can be found by setting y equal to 0. This leaves you with a quadratic equation in one variable, x, and the quadratic formula can be used to solve for x. There might be no solutions, as is the case when the parabola doesn't touch the x-axis. There might be one solution, when the vertex is on the x-axis. Or there might be two solutions, and therefore two horizontal intercepts.
permalinkWhen we know the exact locations of the intercepts (as well as the location of the vertex as found in Section 9.2) then we can plot accurate graphs of quadratic equations.
Subsection 9.5.4 Graphically Solving Equations and Inequalities
permalinkIn Section 9.4, we see how a graph can be used to solve an equation or inequality. Each side of an equation gives you a curve, and where the two curves cross tells you where there are solutions to the equation.
permalinkFor example, to solve the equation x2+xβ1=2x+1, we could plot two curves: y=x2+xβ1 and y=2x+1. We might use a computer to make the graphs for us, as in Figure 9.5.1.
permalinkSince the curves cross at (β1,β1) and (2,5), the solutions are x=β1 and x=2. This means the solution set is {β1,2}.
Exercises 9.5.5 Exercises
Review of Graphing
7.
Find the y-intercept and x-intercept of the line given by the equation. If a particular intercept does not exist, enter none
into all the answer blanks for that row.
x-value | y-value | Location (as an ordered pair) | |
y-intercept | |||
x-intercept |
8.
Find the y-intercept and x-intercept of the line given by the equation. If a particular intercept does not exist, enter none
into all the answer blanks for that row.
x-value | y-value | Location (as an ordered pair) | |
y-intercept | |||
x-intercept |
9.
Find the x- and y-intercepts of the line with equation 5xβ6y=β90. Then find one other point on the line. Use your results to graph the line.
10.
Find the x- and y-intercepts of the line with equation x+5y=β15. Then find one other point on the line. Use your results to graph the line.
Key Features of Quadratic Graphs
11.
Find the axis of symmetry and vertex of the quadratic function.
y=5x2β50x+3
Axis of symmetry:
Vertex:
12.
Find the axis of symmetry and vertex of the quadratic function.
y=β3β30xβ5x2
Axis of symmetry:
Vertex:
13.
Find the axis of symmetry and vertex of the quadratic function.
y=β1βx2+6x
Axis of symmetry:
Vertex:
14.
Find the axis of symmetry and vertex of the quadratic function.
y=β2x2+20x
Axis of symmetry:
Vertex:
15.
Find the axis of symmetry and vertex of the quadratic function.
y=2βx2
Axis of symmetry:
Vertex:
16.
Find the axis of symmetry and vertex of the quadratic function.
y=β2x2β10x+4
Axis of symmetry:
Vertex:
17.
Find the axis of symmetry and vertex of the quadratic function.
y=2x2+10xβ2
Axis of symmetry:
Vertex:
18.
Find the axis of symmetry and vertex of the quadratic function.
y=3x2
Axis of symmetry:
Vertex:
19.
Find the axis of symmetry and vertex of the quadratic function.
y=0.4x2β4
Axis of symmetry:
Vertex:
20.
Find the axis of symmetry and vertex of the quadratic function.
y=5(x+3)2+4
Axis of symmetry:
Vertex:
For the given quadratic equation, find the vertex. Then create a table of ordered pairs centered around the vertex and make a graph.
25.
For y=4x2β8x+5, determine the vertex, create a table of ordered pairs, and then make a graph.
26.
For y=βx2+4x+2, determine the vertex, create a table of ordered pairs, and then make a graph.
27.
For y=x2β5x+3, determine the vertex, create a table of ordered pairs, and then make a graph.
28.
For y=β2x2β5x+6, determine the vertex, create a table of ordered pairs, and then make a graph.
29.
Consider two numbers where one number is 4 less than a second number. Find a pair of such numbers that has the least product possible. One approach is to let x represent the smaller number, and write a formula for a function of x that outputs the product of the two numbers. Then find its vertex and interpret it.
These two numbers are and the least possible product is .
30.
You will build a rectangular sheep enclosure next to a river. There is no need to build a fence along the river, so you only need to build on three sides. You have a total of 420 feet of fence to use. Find the dimensions of the pen such that you can enclose the maximum possible area. One approach is to let x represent the length of fencing that runs perpendicular to the river, and write a formula for a function of x that outputs the area of the enclosure. Then find its vertex and interpret it.
The length of the pen (parallel to the river) should be , the width (perpendicular to the river) should be , and the maximum possible area is .
Graphing Quadratic Equations
31.
Find the y-intercept and any x-intercept(s) of the quadratic curve.
y=x2β2xβ8
y-intercept:
x-intercept(s):
32.
Find the y-intercept and any x-intercept(s) of the quadratic curve.
y=βx2+1
y-intercept:
x-intercept(s):
33.
Find the y-intercept and any x-intercept(s) of the quadratic curve.
y=x2+6x+9
y-intercept:
x-intercept(s):
34.
Find the y-intercept and any x-intercept(s) of the quadratic curve.
y=x2+4x+7
y-intercept:
x-intercept(s):
35.
Find the y-intercept and any x-intercept(s) of the quadratic curve.
y=x2+8x+5
y-intercept:
x-intercept(s):
36.
Find the y-intercept and any x-intercept(s) of the quadratic curve.
y=5x2β8xβ4
y-intercept:
x-intercept(s):
37.
Find the y-intercept and any x-intercept(s) of the quadratic curve.
y=βx+18β5x2
y-intercept:
x-intercept(s):
38.
Find the y-intercept and any x-intercept(s) of the quadratic curve.
y=5xβ2x2
y-intercept:
x-intercept(s):
Graph each curve by algebraically determining its key features.
49.
An object was shot up into the air with an initial vertical speed of 384 feet per second. Its height as time passes can be modeled by the quadratic equation y=β16t2+384t. Here t represents the number of seconds since the objectβs release, and y represents the objectβs height in feet.
After , this object reached its maximum height of .
This object flew for before it landed on the ground.
This object was in the air 3 s after its release.
This object was 704 ft high at two times: once after its release, and again later after its release.
50.
A remote control aircraft will perform a stunt by flying toward the ground and then up. Its height, in feet, can be modeled by the equation h=t2β10t+28, where t is in seconds. The plane
will
will not
51.
An object is launched upward at the height of 280 meters. Its height can be modeled by
where h stands for the objectβs height in meters, and t stands for time passed in seconds since its launch. The objectβs height will be 330 meters twice before it hits the ground. Find how many seconds since the launch would the objectβs height be 330 meters. Round your answers to two decimal places if needed.
The objectβs height would be 330 meters the first time at seconds, and then the second time at seconds.
52.
Currently, an artist can sell 230 paintings every year at the price of $70.00 per painting. Each time he raises the price per painting by $10.00, he sells 10 fewer paintings every year.
Assume he will raise the price per painting x times, then he will sell 230β10x paintings every year at the price of 70+10x dollars. His yearly income can be modeled by the equation:
where i stands for his yearly income in dollars. If the artist wants to earn $21,600.00 per year from selling paintings, what new price should he set?
To earn $21,600.00 per year, the artist could sell his paintings at two different prices. The lower price is per painting, and the higher price is per painting.
Graphically Solving Equations and Inequalities
53.
Use technology to make some graphs and determine how many times the graphs of the following curves cross each other.
y=(286+5x)(78+10x) and y=6000 intersect
zero times
one time
two times
three times
54.
Use technology to make some graphs and determine how many times the graphs of the following curves cross each other.
y=5x3βx2β6x and y=β4x+3 intersect
zero times
one time
two times
three times
55.
Use technology to make some graphs and determine how many times the graphs of the following curves cross each other.
y=0.2(8x2+2) and y=0.2(xβ9) intersect
zero times
one time
two times
three times
56.
Use technology to make some graphs and determine how many times the graphs of the following curves cross each other.
y=1.85(xβ4)2β8.4 and y=x+1 intersect
zero times
one time
two times
three times
57.
The equations y=12x2+2xβ1 and y=5 are plotted.
What are the points of intersection?
Solve 12x2+2xβ1=5.
Solve 12x2+2xβ1>5.
58.
The equations y=βx2+1.5x+5 and y=β5 are plotted.
What are the points of intersection?
Solve βx2+1.5x+5=β5.
Solve βx2+1.5x+5>β5.
59.
The equations y=12x2βxβ1 and y=βx+1 are plotted.
What are the points of intersection?
Solve 12x2βxβ1=βx+1.
Solve 12x2βxβ1>βx+1.
60.
The equations y=14x3 and y=x are plotted.
What are the points of intersection?
Solve 14x3=x.
Solve 14x3>x.
61.
The equations y=βx+4 and y=4x2+x+336 are plotted.
What are the points of intersection?
Solve βx+4=4x2+x+336.
Solve βx+4>4x2+x+336.
62.
The equations y=12x2+2x and y=3β9β2x2+2350xβ5225 are plotted.
What are the points of intersection?
Solve 12x2+2x=3β9β2x2+2350xβ5225.
Solve 12x2+2x>3β9β2x2+2350xβ5225.