please please help!

Question

y=x^2 -4x - 12

a. Minimum or Maximum

b. coordinate of vertex

c.equation of the axis of symmetry

d. y-intercepts

e.x- intercepts

f. domain

g. range

Answer 1

The quadratic function y = x² - 4x - 12

The quadratic function represent a parabola.

y = ax² + bx + c

a = 1, b = - 4, c = 12

a) Since a = 1  is positive number the parabola opens up and has minimum value.

Vertex is minimum point.

 

c) Axis of symmetry x = -b/2a

x = -(-4)/2(1)

x = 4/2

x = 2

Equation of axis of symmetry x = 2.

Substitute x = 2 in y = x² - 4x - 12.

y = 2² - 4(2) - 12

y = 4 - 8 - 12

y = - 16

b) Coordinate of vertex (x, y) = (2, - 16)

 

d) To find y intercepts substitute x = 0 in y = x² - 4x - 12.

y = 0² - 4(0) - 12

y = - 12

y intercept is - 12.

 

e) To find x intercepts substitute y = 0 in y = x² - 4x - 12.

0 = x² - 4x - 12

x² - 6x + 2x - 12 = 0

x( x - 6) + 2(x - 6) = 0

(x - 6)( x + 2) = 0

x - 6 = 0 and x + 2 = 0

x = 6 and x = - 2

x intercepts are 6 and - 2.

 

f) Domain of y = x² - 4x - 12 is all real numbers.

 

g) In the minimum point y  = - 16  so the graph of parabola cannot be lower than - 16.

Thus the range of function y  ≥ - 16

Range of the function is  {y |y  ≥ - 16}.