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.
To find the number of points of intersections, we will draw the graphs of both functions on the same coordinate grid. Let's start with the parabola.
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 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
| x=- 3± 1 | |
|---|---|
| x_1=- 3+ 1 | x_2=- 3- 1 |
| x_1=- 2 | x_2=- 4 |
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