Andrew M. answered 05/26/16
Tutor
New to Wyzant
Mathematics - Algebra a Specialty / F.I.T. Grad - B.S. w/Honors
f(x) = (x+2)(5x-25)
We can factor a 5 out of (5x-25) giving:
f(x) = 5(x+2)(x-5)
replace f(x) with y
y = 5(x+2)(x-5)
a) The x-intercepts are where y=0
5(x+2)(x-5) = 0 at x=-2 and x=5
x - intercepts are (-2,0) and (5,0)
b) y intercept is where x=0
5(0+2)(0-5) = 5(2)(-5) = -50
y intercept is at (0, -50)
c) Find the vertex
the vertex of a quadratic y =ax2+bx+c
is at (b/2a, f(-b/2a))
y = 5(x+2)(x-5)
y = 5(x2-3x-10)
y = 5x2-15x-50
a = 5, b=-15, c=-50
-b/2a = 15/10 = 3/2 or 1 1/2
f(3/2) = 5(3/2)2-15(3/2)-50
= 5(9/4) - 45/2 - 50
= -45/4 - 200/4
= -245/4
= -61 1/4
Vertex of the parabola is at
(3/2 , -245/4)
d) Since the coefficient of the squared term in
y= 5x2-15x-50 is positive, the parabola opens upward.
Note: we can also tell this because the vertex is below the
x-axis and there are real x-intercepts at (-2,0) and (5,0)
e) The equation for the line of symmetry is the line through the
x coordinate of the vertex because the square term is the x.
x = 3/2 is the equation for the line of symmetry.