3.1 6, a quadratic function is given

Question

A quadratic function f is given.

f(x) = −4x² − 20x + 4

(a)Express f in vertex form.

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

vertex:

X intercept (smaller x value):

X intercept (larger x value)

Y intercept:

c. find the domain and range of f in interval notation

Answer 1

Although the use of the complete-the-square method is most common in this type of problem, you can also use the following method:

In standard form, the x-value of the vertex (h) is found by x = -b/2a, and then the y-value of the vertex (k) is found by subbing that back into the equation. Here's what that looks like for your example:

y = -4x² - 20x + 4

Thus, h = -(-20)/(2•-4) = -20/8 = -5/2 or -2.5

Subbing this x-value in gives

k = -4(-2.5)² - 20(-2.5) + 4 = 29

Finally, since the a-value in vertex form is the same a-value in standard form, a = -4. This leaves you with the following vertex form: y = -4(x + 2.5)² + 29.

For the characteristics:

The vertex is (h,k) so it is (-2.5, 29)

The y-intercept can be taken directly from the given standard form equation (c), so it is 4.

The x-intercepts can be found by setting either form of the equation equal to zero and solving for x (use quadratic formula for the standard form and use square roots for the vertex form).

Domain and Range: The domain of any unrestricted quadratic is "all real numbers," so it becomes this in interval notation: (-∞, ∞)

The range is always either y ≥ k or y ≤ k. Since you have negative a-value, your parabola opens downward, so your range is y ≤ 29. In interval notation, this becomes (-∞, 29]

Answer 2

f(x)=-4x^2 -20x +4

= -4(x^2+5x) +4

= -4(x^2+5x +25/4) + 4-25

= -4(x+2.5)^2 - 21

= -4(x^2+5x+1)

x=- 5/2+/-.5sqr(25-4) = -2.5 +/-.5sqr21

vertex = (-2.5, f(2.5)) = (-2.5, 29)

x intercepts =-2.5+/-.5sqr21

y intercept =(0,4)

domain = all real numbers

range y</=29

Answer 3

f(x) = -4(x² + 5x) + 4 = -4(x² + 5x + 25/4 - 25/4) + 4 = -4(x² + 5x + 25/4) + 29 = -4(x + 5/2)² + 29

Vertex = (-5/2, 29)

Domain = (-∞, ∞)

Range : Since the coefficient of x² is negative, the parabola opens downward with highest point (-5/2, 29).

Range = (-∞, 29)

y-intercept: (0, 4)

x-intercepts: -4(x + 5/2)² + 29 = 0. Solve for x.