Problema Solution
Graph y=4x2(squared)-6x-2 on the standard axes. Find the x-intercepts, y-intercept, and vertex. Determine whether the vertex is a relative maximum or a relative minimum.
Answer provided by our tutors
the x-intercepts are when y = 0
y = 4x2 - 6x - 2
0 = 4x2 - 6x - 2
x = 3 +/- sqrt(17) / 4
so ((3+sqrt(17))/4,0) and ((3-sqrt(17))/4,0) are the x-intercepts
y-intercept is when x = 0
y = 4(0)2 - 6(0) - 2
y = -2
so (0,-2) is the y-intercept
vertex:
x = -b/2a
x = 6/8
x = 3/4
y = 4(3/4)2 - 6(3/4) - 2
y = -17/4
(3/4,-17/4) is the vertex and it is a minimum since the coefficient of the squared term is positive & the parabola opens up