Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
2. Section 6.2
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Exercise 75 Page 347

Practice makes perfect
a Let's have a look at the parent graph to all quadratics, y=x^2.

As we can see, the graph of y=x^2 intersects the x-axis once at the origin which means this solution is a double root. Now, let's think about the function y=x^2-6. This is a vertical translation of y=x^2 by 6 units down.

As we can see, the given function intersects the x-axis twice, which means it has two real roots.

b From Part A we know that the parent function to quadratics has a double root at the origin. The function y=x^2+6 shows a vertical translation of the parent function by 6 units up.

As we can see, this graph does not intersect the x-axis at all, which means it does not have any real roots, so it must have complex roots.

c We can use the discriminant in the Quadratic Formula to tell whether the function has real roots or not.
x=b ±sqrt(b^2-4ac)/2aThe following applies for the number of roots in a quadratic equation. Complex Roots: b^2-4ac<0 One Real Root: b^2-4ac=0 Two Real Roots: b^2-4ac>0 In our equation, we see that a= 1, b= - 2 and c= 10. By substituting these values into the discriminant, we can determine if the function has real or complex roots.
b^2-4ac
( - 2)^2-4( 1)( 10)
â–Ľ
Simplify
(- 2)^2-40
2^2-40
4-40
- 36
Since the discriminant is negative, the function has complex roots.
d Like in Part C, we will substitute the values of a, b and c into the discriminant. Examining the equation, we see that a= 1, b= - 2 and c= - 10.
b^2-4ac
( - 2)^2-4( 1)( - 10)
â–Ľ
Simplify
(- 2)^2+40
2^2+40
4+40
44
Since the discriminant is positive, the equation has two real solutions.
e Examining the equation, we notice that it is written in graphing form. In this form, the vertex is easy to identify.
Graphing Form:& y=a(x- h)^2+ k Vertex:& ( h, k)

Notice that the value of a determines whether the vertex is a maximum or a minimum. A positive a gives a minimum, and a negative a gives a maximum. Examining the equation, we can identify the vertex and what type of vertex it is. Graphing Form:& y=(x- 3)^2+( -4) Vertex:& ( 3, - 4) Type of vertex:& a>0 → minimum Since the function's vertex is a minimum at (3,- 4), which is a point that is below the x-axis, the graph must have two real roots.

f Like in Part E, we have a quadratic function written in graphing form.

graphing form:& y=(x- 3)^2+ 4 vertex:& ( 3, 4) type of vertex:& a>0 → minimum Since the function's vertex is a minimum at (3,4), which is a point that is above the x-axis, the graph has complex roots.