Big Ideas Math Algebra 1, 2015
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Big Ideas Math Algebra 1, 2015 View details
5. Using Intercept Form
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Exercise 76 Page 456

Recall the general factored form of a cubic function and use the given points to find its equation.

Sample Answer: f(x)=- x^3+6x^2-11x+6

Practice makes perfect

We want to find a cubic function that satisfies the given conditions. Let's think about what the given information tells us.

  • A cubic function is a 3-degree function — this degree tells us that the function can have one or three real zeros.
  • The function passes through (0,6) — this tells us the location of the y-intercept
Let's notice that we do not have any information about the zeros, so we can choose any values we want, except for 0, which is already a y-intercept, and then find such a that the function will be passing through the given point. Let's recall the general form of a factored cubic equation. f(x)=a(x-p)(x-q)(x-r)In the above equation, p, q, and r are the x-intercepts of the function. For simplicity we will let p= 1, q= 2 and r=3. f(x)=a(x- 1)(x- 2)(x-3) Now, since the function passes through the point (0,6), we can substitute 0 for x and 6 for f(x) in our partial equation, and solve for a.
f(x)=a(x-1)(x-2)(x-3)
6=a( 0-1)( 0-2)( 0-3)
â–Ľ
Solve for a
6=a(- 1)(- 2)(- 3)
6=- 6a
- 1=a
a=- 1
Knowing that a=- 1 we can write the equation of our cubic function. f(x)=- 1(x-1)(x-2)(x-3) ⇕ f(x)=- (x-1)(x-2)(x-3) Finally, let's rewrite the function in order to obtain its standard form.
f(x)=- (x-1)(x-2)(x-3)
f(x)=(1-x)(x-2)(x-3)
â–Ľ
Multiply parentheses
f(x)=(1(x-2)-x(x-2))(x-3)
f(x)=(x-2-x(x-2))(x-3)
f(x)=(x-2-x^2+2x)(x-3)
f(x)=(3x-2-x^2)(x-3)
f(x)=3x(x-3)-2(x-3)-x^2(x-3)
f(x)=3x^2-9x-2(x-3)-x^2(x-3)
f(x)=3x^2-9x-2x+6-x^2(x-3)
f(x)=3x^2-9x-2x+6-x^3+3x^2
f(x)=6x^2-11x+6-x^3
f(x)=- x^3+6x^2-11x+6
Note that this is only one of the infinitely many functions with the given characteristics. Any function passing through (0,6) will be correct.