Function problem

Question

Consider the function: f(x) = 3x^2 + 36x + 60

The y-intercept is at y= ___________

The x-intercepts are at x= _____________

(Enter solutions separated by a comma)

The vertex is at the point _________

Answer 1

Function problem: Consider the function: f(x) = 3x² + 36x + 60. The y-intercept is at y = 60. The x-intercepts are at x = – 2 and x = – 10. The vertex is at the point (x, y) = (h, k) = (– 6, – 48).

Detailed Solution:

Given/Known: f(x) = 3x² + 36x + 60 = 3(x² + 12x + 20)

f(x) = a(x – h)² + k <== write in vertex form

(b/2)² = (12/2)² = (6)² = 36

f(x) = 3(x² + 12x + 36 – 36 + 20)

f(x) = 3((x + 6)² – 16)

f(x) = 3x² + 36x + 60 = 3(x² + 12x + 20) = 3(x + 6)² – 48 <== in vertex form

f(x) = 3(x + 6)² – 48 <== in vertex form

a) The y-intercept: y = f(0) = 3(0 + 6)² – 48 = 3(6)² – 48 = 3(36) – 48 = 108 – 48 = 60

The y-intercept is at y = 60

a) The x-intercept: f(x) = 0 ==> 3(x + 6)² – 48 = 0 ==> (x + 6)² – 16 = 0 ==> (x + 6)² = 16 ==> x + 6 = ±4 ==> x = – 6 ± 4. Therefore: x = – 6 + 4 and x = – 6 – 4. x = – 2 and x = – 10.

The x-intercepts are at x = – 2 and x = – 10

c) The vertex is: f(x) = a(x – h)² + k ==> f(x) = 3(x + 6)² – 48 <== in vertex form. Therefore: a = 3, x = h = – 6 , and y = k = – 48

The vertex is at the point: (x, y) = (h, k) = (– 6, – 48)

Answer 2

Recall the y-intercept occurs when x=0 and the x-intercept occurs when y=0, where, when considering a function f, y=f(x)=3x²+36x+60. So to find the x-intercept, we must solve the equation

0 = 3x²+36x+60

As we are considering a quadratic equation, there are a couple approaches we could employ: factoring or using the quadratic formula. A first glance, this seems like it will be an easy factorization.

0=3x²+36x+60 = 3(x²+12x+20)=3(x+2)(x+10)

0=(x+2)(x+10)

Using the fact that if a·b=0, then either a=0 or b=0, we can conclude either

x+2=0 or x+10 ⇒ x=-2 or -10

Hence, there are two x-intercepts!

To find the vertex, it helps to know that every parabola (ie, quadratic) can be written in the form

y=a(x-h)²+k, where (h,k) is the vertex of the parabola.

We can convert any quadratic of the form ax²+bx+c to the above form by completing the square:

1. First, factor out the leading coefficient: 3(x²+12x+20)
2. Take the coefficient of x, divide it by two, square it, and then simultaneously add and subtract it the result from inside the parentheses: 3(x²+12x+36-36+20)
3. Distribute the 3, but keep the first three terms in parentheses: 3(x²+12x+36)+3(20-36)
4. Observe the quadratic x²+12x+36 is a perfect square and can be factored to (x+6)²
5. Simplify the constant terms, and we have the result: y=3(x+6)²-48. However, it does not quite match the form y=a(x-h)²+k. We need to make a subtle change in representation, 3(x- (-6))² + -48. Thus the vertex is (-6,-48)