Factor the expression in the function y = x^2 - 3x - 4. use the result to identify the zeros of the function, the maximum or minimum of the function, and the axis of symmetry of its graph.
* with this, its due tomorrow! *
*i know that after you factor the equation, it becomes (x-4)(x+1)=0, and the zeros are 4 or -1, but i really don't get how to find the maximum or minimum or the axis of symmetry with the zeros.
*posted twice on accident*

y = (x - 4)(x + 1) ; zeroes at (-1, 0) and (4, 0)

minimum at (3/2, -25/4); axis of symmetry at x = 3/2

Step-by-step explanation:

Your equation is: y = x² - 3x - 4

1. Factor the quadratic

x² - 3x - 4 = (x - 4)(x + 1)

2. Solve the quadratic

x - 4 = 0     x + 1 = 0

    x = 4           x = -1

The zeroes are at (-1, 0) and (4, 0).

3. Find the vertex

The standard form for the equation of a parabola is

y = ax² + bx + c

Your equation is

y = x² - 3x - 4

a = 1, b = -3, c = -4

The vertex form of a parabola is

y = a(x - h)² + k

where (h, k) is the vertex of the parabola.

h = -b/(2a) and k = f(h)

a is positive, so the parabola opens upward, and the vertex is a minimum.

h = -(-3)/(2×1)

  =        3/2

k = 1(3/2)² - 3(3/2) - 4

  =   9/4    -    9/2    - 4

  =   (9 - 18 - 16)/4

  = -25/4

h = 3/2; k = -25/4

The vertex is at (3/2, -25/4).

The axis of symmetry is a vertical line passing through the vertex at x = 3/2.

The Figure below shows that your equation us an upward pointing parabola with zeroes at (-1,0) and (4, 0), vertex at (1.5, -6.25), and axis of symmetry at x = 1.5.

-8/2y-8=5/y+4-7y+8/y^2-16

we move all terms to the left:

-8/2y-8-(5/y+4-7y+8/y^2-16)=0

we can not solve this equation