Chapter Closure
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Exercise 142 Page 576
a
To find the number of points of intersection, draw the graphs of both functions on the same coordinate grid.
b
To solve the equation ax^2+bx+c=0, use the Quadratic Formula.
a
Two points of intersection
b
(- 2, 0), (- 4,4)
a
We want to determine the number of points of intersection of the given system of equations. Note, that the first equation of the system is a quadratic equation and the second one is a linear equation.
y=(x+2)^2 & (I) y=- 2x-4 & (II) In other words, we want to find the number of points in which the graphs of the given equations — a parabola and a line — intersect. When it comes to a line and a parabola, we have three possibilities.
Graphing the Parabola
To graph the parabola, let's recall the graphing form of a quadratic function. y=a(x- h)^2+ k In this form the vertex of the parabola is the point ( h, k), and the axis of symmetry is the vertical line x= h. Now consider the given function. y=(x+2)^2 ⇕ y=1(x-( - 2))^2 + 0 We can see that h= - 2 and that k= 0. Therefore, the vertex is ( - 2, 0) and the axis of symmetry is x= - 2. To graph the function we will make a table of values. Make sure to include x-values to the left and to the right of the axis of symmetry.
| x | (x+2)^2 | y=(x+2)^2 |
|---|---|---|
| - 4 | ( - 4+2)^2 | 4 |
| - 3 | ( - 3+2)^2 | 1 |
| - 1 | ( - 1+2)^2 | 1 |
| 0 | ( 0+2)^2 | 4 |
Let's now draw the axis of symmetry x=- 2 and the parabola that connects the obtained points and the vertex.
Graphing the Line
We will graph the linear function on the same coordinate plane. For a linear equation written in slope-intercept form, we can identify its slope m and y-intercept b. y= - 2x+( - 4) The slope of the line is - 2 and the y-intercept is - 4.
Points of Intersections
Finally, we can determine the number of points of intersections.
As we can see, the graphs of the given functions have two points of intersection.
b
We will solve the given system of equations using the Substitution Method.
y=(x+2)^2 & (I) y=- 2x-4 & (II) Notice that y-variable is isolated in both equations. Since the expression equal to y in Equation (II) is simpler, we will substitute its value - 2x-4 for y in Equation (I).
y=(x+2)^2 y=- 2x-4
Substitute
(I): y= - 2x-4
- 2x-4=(x+2)^2 y=- 2x-4
▼
(I): Simplify
ExpandPosPerfectSquare
(I): (a+b)^2=a^2+2ab+b^2
- 2x-4=x^2+4x+4 y=- 2x-4
AddEqn
(I): LHS+2x=RHS+2x
- 4=x^2+6x+4 y=- 2x-4
AddEqn
(I): LHS+4=RHS+4
0=x^2+6x+8 y=- 2x-4
RearrangeEqn
(I): Rearrange equation
x^2+6x+8=0 y=- 2x-4
In Equation (I) we have a quadratic equation in terms of only the x-variable.
This result tells us that we have two solutions for x. One of them will use the positive sign and the other one will use the negative sign.
| x=- 3± 1 | |
|---|---|
| x_1=- 3+ 1 | x_2=- 3- 1 |
| x_1=- 2 | x_2=- 4 |
Now, consider Equation (II). y=- 2x-4 We can substitute x=- 2 and x=- 4 into the above equation to find the values for y. Let's start with x=- 2.
We found that y=0 when x=- 2. One solution of the system, which is a point of intersection of the parabola and the line, is (- 2,0). To find the other solution we will substitute - 4 for x in Equation (II) again.
We found that y=4 when x=- 4. Therefore our second solution, which is the other point of intersection of the parabola and the line, is (- 4,4). Thus, our answer in Part A was correct — we have two points of intersection, which are (- 2, 0) and (- 4, 4).
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