Question 143459
Set y=0 :

x^2-5x-4=0

Use the quadratic formula to find where y is zero (the x-intercepts):
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

*[invoke quadratic "x", 1, -5, -4 ]

To find the y intercept we set x=0;

0^2-5(0)-4=y
It follows that y=-4. Thus, (0,-4) is the y-intercept.

The vertex is x=-b/(2a) when the equation is in form ax^2+bx+c=0. Thus, 5/2=x is the vertex. Now, find the corresponding y coordinate for x=5/2. 

(5/2)^2-5(5/2)-4=25/4-50/4-16/4=-41/4
Thus, the vertex is at (5/2,-41/4). The graph is above.