Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
3. Section 9.3
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Exercise 86 Page 513

Find the vertex and the axis of symmetry of the parabola.

Graph:

Points of Intersection: Approximately (- 3.5,- 7.5) and (0.5, - 3.5)
Number of Intersections: 2

Practice makes perfect

To solve the system of equations by graphing, we will draw the graph of the quadratic function and the linear function on the same coordinate grid. Let's start with the parabola.

Graphing the Parabola

To graph the parabola, we first need to identify a, b, and c. y=x^2-2x+2 ⇔ y= 1x^2+( - 2)x+ 2 For this equation we have that a= 1, b= - 2, and c= 2. Now, we can find the vertex using its formula. To do this, we will need to think of y as a function of x, y=f(x). Vertex of a Parabola: ( - b/2 a,f(- b/2 a) ) Let's find the x-coordinate of the vertex.
- b/2a
- - 2/2( 1)
- - 2/2
-2/-2
1
We use the x-coordinate of the vertex to find its y-coordinate by substituting it into the given equation.
y=x^2-2x+2
y= 1^2-2( 1)+2
y=1-2+2
y=1
The y-coordinate of the vertex is 1. Thus, the vertex is at the point (1,1). With this, we also know that the axis of symmetry of the parabola is the line x=1. Next, let's find two more points on the curve, one on each side of the axis of symmetry.
x x^2-2x+2 y=x^2-2x+2
^2-2( )+2 2
2 2^2-2(2)+2 2

Both ( ,2) and (2,2) are on the graph. Let's form the parabola by connecting these points and the vertex with a smooth curve.

Graphing the Line

Let's now 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=4x-7 ⇔ y=4x+ 7 The slope of the line is 4 and the y-intercept is 7.

Finding the Solutions

Finally, let's try to identify the coordinates of the points of intersection of the parabola and the line.

It looks like the points of intersection occur at (- 0.75, 4) and (6.75,34).

Checking the Answer

To check our answers, we will substitute the values of the points of intersection in both equations of the system. If they produce true statements, our solution is correct. Let's start with (- 0.75, 4).
y=4x+7 & (I) y=x^2-2x+2 & (II)

(I), (II): x= - 0.75, y= 4

4? =4( -0.75)+7 4? =( -0.75)^2-2( -0.75)+2
Simplify right-hand side
4? =4(-0.75)+7 4? =0.5625-2(-0.75)+2

(I), (II): Multiply

(I): - a(- b)=a* b

4? =-3+7 4? =0.5625+1.5+2

(I), (II): Add terms
4=4 ✓ 4≈ 4.0625 ✓
Equation (I) produced a true statement. In Equation (II), the answer is an approximation. This is because we could not state an exact answer just by looking at the graph, but we obtained a decent approximation. Therefore, we can say that (- 0.75, 4) is an approximated answer. Let's continue by checking (6.75, 34).
y=4x+7 & (I) y=x^2-2x+2 & (II)

(I), (II): x= 6.75, y= 34

34? =4( 6.75)+7 34? =( 6.75)^2-2( 6.75)+2
Simplify right-hand side
34? =4(6.75)+7 34? =45.5625-2(6.75)+2

(I), (II): Multiply

(I): - a(- b)=a* b

34? =27+7 34? =45.5625-13.5+2

(I), (II): Add terms
34=34 ✓ 34≈ 34.0625 ✓
Equation (I) produced a true statement. Just like before, in Equation (II) the answer is an approximation. This is because we could not state an exact answer just by looking at the graph, but we obtained a decent approximation. Therefore, we can say that (6.75, 34) is also an approximated answer.