please please help!

Question

y=-3(x-2)^2-4

a.minimum of maximum

b.coordinate of the vertex

c.equation of the axis of symmetry

d x intercept

e. y intercepts

Answer 1

The function y = - 3(x - 2)² - 4

y =  - 3(x²  + 4 - 4x) - 4

y = - 3x² - 12 + 12x - 4

y = - 3x²  + 12x - 16

The quadratic function y = - 3x² + 12x - 16

The quadratic function represent a parabola.

y = ax² + bx + c

a = - 3, b = 12, c = - 16

a) Since a = - 3  is negative number the parabola opens down and has maximum value.

Vertex is maximum point.

 

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

x = -(12)/2(-3)

x = 2

Equation of axis of symmetry x

= 2.

Substitute x = 2 in y = - 3x² + 12x - 16.

y = - 3(2)² + 12(2) - 16

y = - 12 + 24 - 16

y = - 4

b) Coordinate of vertex (x, y) = (2, - 4)

 

d) To find x intercepts substitute y = 0 in y = - 3x² + 12x - 16.

0 = - 3x² + 12x - 16

b² - 4ac = (12)² - 4(-3)(-16)

= 144 - 192 = -48

The discriminant is negative then the roots are imaginary.

There is no x intercepts.

 

e) To find y intercepts substitute x = 0 in y = - 3x² + 12x - 16.

y = - 3(0)² + 12(0) - 16

y = - 16

y intercept is - 16.