Question

A quadratic function f is given.

f(x) = x² + 6x + 8

(a) Express f in standard form.

f(x)

=

(b) Find the vertex and x- and y-intercepts of f. (If an answer does not exist, enter DNE.)

vertex	(x, y)	=
x-intercepts	(x, y)	=	(smaller x-value)
	(x, y)	=	(larger x-value)
y-intercept	(x, y)	=

(d) Find the domain and range of f. (Enter your answers using interval notation.)

domain
range

Answer 1

(a) f(x) = (x + 3)^2 - 1

(b) Vertex : (-3 , -1)

Smaller x-value intercept : (-4 , 0)

Larger x-value intercept : (-2 , 0)

y-intercept : (0 , 8)

(c) Domain : (-∞ , ∞)

Range : [-1 , ∞)

(a) f(x) = x^2 + 6x + 8

f(x) = a (x - h)^ + k

f(x) - 8 = x^2 + 6x

On completing the square, f(x) - 8 + 9 = x^2 + 6x + 9

f(x) + 1 = (x + 3)^2

f(x) = (x + 3)^2 - 1

(b) f(x) = a (x - h)^ + k

Vertex : (h , k)

f(x) = (x + 3)^2 - 1

Vertex : (-3 , -1)

To find x-intercepts, solve for f(x) = 0.

(x + 3)^2 - 1 = 0

(x + 3)^2 = 1

x + 3 = ± 1

x = -3 + 1 = -2 or x = -3 - 1 = -4

x-intercepts are(-4,0) and (-2,0).

Smaller x-value intercept : (-4 , 0)

Larger x-value intercept : (-2 , 0)

To find the y-intercept, plug in x = 0 i.e. find f(0).

f(x) = x^2 + 6x + 8

f(0) = 0^2 + 6(0) + 8 = 8

y-intercept : (0 , 8)

(c) For f(x) = a (x - h)^ + k, domain is all real numbers and range depends on k. As the value of a is positive, graph opens up, the lowest point is the vertex and so range is y ≥ k.

f(x) = (x + 3)^2 - 1

Domain : (-∞ , ∞)

Range : [-1 , ∞)