y = x^2-13x+36
clearly the y-intercept is (0,36)
y = x^2 - 13x + (13/2)^2 + 36 - (13/2)^2
y = (x - 13/2)^2 - 25/4
So, the vertex is at (13/2,-25/4)
y = x^2-13x+36
y = (x-4)(x-9)
The x-intercepts are at (0,4),(0,9)
The coefficient of x^2 is positive, so it opens up.
(C) No idea what you mean by the size of a parabola. Stupid question.
x^2 = 4py
has its focus at y=p
So, for 4y = x^2, the focus is at (0,1/4)
We have shifted the parabola right by 13/2 and down by 25/4, so the focus is at (0+13/2, 1/4-25/4) = (13/2,-6)
http://www.wolframalpha.com/input/?i=x^2%E2%88%9213x%2B36%E2%89%A40
take the indicated quadratic inequality x^2−13x+36≤0
A)determine the type of graph
B)does it go up or down
C) the size
D)graph the equation
E)define the vertex
F)X -intercept
G) Y-intercept
H)focal point
1 answer