Quartics using P(X) = af(b(x-c)+d
Recall Q(X) = (x-2)(x-1)(x+1)(x+2)
a)Choosing a
(i) Rewrite Q(x) for a. Let b = 1, c = 0 and d = 0
(ii) Make 3 copies of the same function on your calculator
Let a= -1, 1 and 2
Your equations are
Y1= __________(a=-1)
Y2= __________(a=1)
Y3= __________(a=2)
(iii) The above functions share the same x - intercepts and are called a family of functions. Find the value of a which causes Q(X) to pass through
(I)(0,4) a = _______
(II) (3,80) a = _______
a) Choosing a:
(i) Rewrite Q(x) for a. Let b = 1, c = 0, and d = 0
Q(x) = (x-2)(x-1)(x+1)(x+2)
= (x^2 - 4)(x^2 - 1)
= x^4 - 5x^2 + 4
P(x) = a * f(b(x-c) + d)
P(x) = a * f(x)
(ii) Make 3 copies of the same function on your calculator
Let a = -1, 1, and 2. Your equations are:
Y1 = -f(x)
Y2 = f(x)
Y3 = 2f(x)
(iii) The above functions share the same x-intercepts and are called a family of functions. Find the value of a which causes Q(x) to pass through:
(I) (0,4) a = _______
To find the value of a that causes Q(x) to pass through (0,4), we substitute (0,4) into P(x) = a * f(x) and solve for a.
P(0) = a * f(0)
4 = a * f(0)
Since f(0) = Q(0) = 0^4 - 5(0)^2 + 4 = 4, we have:
4 = a * 4
a = 1
Therefore, a = 1 causes Q(x) to pass through (0,4).
(II) (3,80) a = _______
To find the value of a that causes Q(x) to pass through (3,80), we substitute (3,80) into P(x) = a * f(x) and solve for a.
P(3) = a * f(3)
80 = a * f(3)
Since f(3) = Q(3) = 3^4 - 5(3)^2 + 4 = 80, we have:
80 = a * 80
a = 1
Therefore, a = 1 causes Q(x) to pass through (3,80).
a) Choosing a
(i) To rewrite Q(x) for a, we substitute b = 1, c = 0, and d = 0 into the given equation P(x) = af(b(x-c)+d). Since Q(x) is given as (x-2)(x-1)(x+1)(x+2), we have:
Q(x) = a(1(x-0)+0) * (1(x-1)+0) * (1(x+1)+0) * (1(x+2)+0)
Simplifying this expression, we get:
Q(x) = a(x-2)(x-1)(x+1)(x+2)
(ii) To make 3 copies of the same function on your calculator with different values of a, let's substitute a = -1, 1, and 2 into the equation Q(x) = a(x-2)(x-1)(x+1)(x+2).
Your equations will be:
Y1 = -1(x-2)(x-1)(x+1)(x+2)
Y2 = 1(x-2)(x-1)(x+1)(x+2)
Y3 = 2(x-2)(x-1)(x+1)(x+2)
(iii) To find the value of a which causes Q(x) to pass through the given points (0,4) and (3,80), you can substitute the x and y coordinates into each of the three equations and solve for a.
For the point (0,4), we have:
Y1(0) = -1(0-2)(0-1)(0+1)(0+2) = 4
Y2(0) = 1(0-2)(0-1)(0+1)(0+2) = -4
Y3(0) = 2(0-2)(0-1)(0+1)(0+2) = -16
Only Y1 gives us the correct y-value of 4. Therefore, for point (0,4), we have a = -1.
For the point (3,80), we have:
Y1(3) = -1(3-2)(3-1)(3+1)(3+2) = 20
Y2(3) = 1(3-2)(3-1)(3+1)(3+2) = 120
Y3(3) = 2(3-2)(3-1)(3+1)(3+2) = 240
Only Y3 gives us the correct y-value of 80. Therefore, for point (3,80), we have a = 2.
To summarize:
For point (0,4), a = -1
For point (3,80), a = 2