f Begin by factoring out - 1. Can you now factor the equation?
G
g Press the Y= button and write the equation on the first row.
H
h Think about perfect square trinomials.
A
a 2
B
b 1
C
c 1
D
d 2
E
e 2
F
f 1
G
g See solution.
H
h See solution.
Practice makes perfect
a Examining the equation, we see that the right-hand side is written in factored form. This means, if we equate y with 0, we can solve for the x-intercept(s) by setting each factor equal to 0 and solving for x. Since the factors are different, there will be two x-intercepts.
b In this case, the right-hand side is a squared binomial. Let's rewrite it in factored form.
y=(x+1)^2 ⇔ y=(x+1)(x+1)
Like in Part A, we should set y equal to 0 and solve for x by using the Zero Product Property. However, since both factors are the same, the x-intercepts will be the same. Therefore, there is only one x-intercept.
c It is not perfectly clear how many intercepts there are since the right-hand side is not written in factored form. However, examining the right-hand side, we see that the expression is a perfect square trinomial.
cclcc
a^2 & -7pt + -7pt & 2(a)(b) & -7pt + -7pt & b^2
x^2 & -7pt + -7pt & 2(x)(3) & -7pt + -7pt & 3^2
This means we can rewrite the right-hand side in the format (a+b)^2.
y=x^2+2(x)(3)+3^2
⇕
y=(x+3)^2
Similar to Part B, we have a squared binomial on the right-hand side which means the number of x-intercepts is one.
d In this case, we do not have any easy way of rewriting the right-hand side in factored form. However, if we complete the square, we can rewrite the right-hand side expression in graphing form which allows us to identify the coordinate of its vertex.
Graphing Form:& y=a(x-h)^2+k
Vertex:& (h,k)To complete the square, we have to add the square of half the coefficient to x to both sides of the equation.
Having rewritten the equation, we can identify the vertex.
Graphing Form:& y=(x-7/2)^2-9/4
Vertex:& (7/2,- 9/4)
As we can see,the vertex has a negative y-coordinate which means it's below the x-axis. Since the coefficient to x^2 is positive, the parabola will open upwards and therefore, it will intersect the x-axis twice.
e Like in Part D, we must rewrite the function into graphing form by completing the square. This will allow us to locate its vertex.
Having rewritten the equation, we can identify the vertex.
Graphing Form:& y=(x-3)^2-1
Vertex:& (3,-1)
As we can see, the vertex has a negative y-coordinate which means it's below the x-axis. Since the coefficient to x^2 is positive, the parabola will open upwards and therefore, it will intersect the x-axis twice.
f Examining the right-hand side of the equation, the first thing we notice is that all terms are negative. This means we can start by factoring out -1.
Having factored out -1, we notice that the expression within the parentheses is a perfect square trinomial.
cclcc
a^2 & -7pt + -7pt & 2(a)(b) & -7pt + -7pt & b^2
x^2 & -7pt + -7pt & 2(x)(2) & -7pt + -7pt & 2^2
This means we can rewrite the expression that's inside the parentheses in the format (a+b)^2.
y=- (x^2+4x+4)
⇕
y=- (x+2)^2
Similar to Parts B and C, we have a squared binomial on the right-hand side which means the number of x-intercepts is one.
g Let's draw the first equation on the calculator. Press the Y= button and write the equation on the first row.
By pressing GRAPH we draw the equation's graph.
As we can see, the parabola dips just below the x-axis which gives it two solutions. To make sure this is the case, we can reset the window to zoom it in.
We were right. If we follow the same procedure for the remaining functions, we can check them as well.
h From previous parts, we notice that all parabolas who touch the x-axis can be written in the following format. y=(x+ a)^2 ⇔ y=(x+a)(x+a) [0.5em]
or [0.5em]
y=(x-a)^2 ⇔ y=(x-a)(x-a)
Therefore, what they all have in common is that the equation of the parabola can be written as a product of two identical binomials.
