If you let eb 2x = t, then you have t2 = (eb 2x)2 = e2⋅ b 2x = ebx So, the equation becomes at2 + ct + d, which you can indeed solve with the quadratic equation. Then, assuming you found two …

The graph is inserted. graph {y>=2x^2-2x-5 [-6, 6, -12, 12]} y = 2x^2-2x-5=2 (x-1/2)^2-11/2, giving the standard form (x-1/2)^2=4 (1/8) (y+11/2) The vertex ( least-y point ) of this parabola is (1/2, …

#"then the factors are" (x-a), (x-b)" and " (x-c)# #"the polynomial is the product of the factors"# #rArrp (x)=k (x-a) (x-b) (x-c)larrcolor (blue)"k is a multiplier"# #"identify the roots from the …

See below The idea is that you start at time t=0 at the base of the Ferris wheel, and for practical reasons say that the radius is r=1, so you start at the height y=-1. Then, since you're taking …

Oct 23, 2016 · Well, all you have to do is to plug two values for x and find the values of y then graph the line. It will surely pass through the origin. Let's say we choose the values 1 and 2 for …

See a solution process below: First, solve for two points which solve the equation and plot these points: First Point: For x = 0 y = (-1/2 * 0) + 5 y = 0 + 5 y = 5 or (0, 5) Second Point: For x = 4 y …

Slant asymptote: 2x+y-4=0 Vertical asymptote { uarr x = -2 darr . The asymptotes design proximity ( getting the graph closer ), as the end behavior. x-intercept:2sqrt2. y-intercept: 8. By actual …

How do you use the important points to sketch the graph of #y=3x^2+6x-1 #?

Feb 1, 2016 · The graph has a V shape with a vertex at (-2,0) a slope for the left arm of (-1) and a slope for the right arm of (+1) abs (x+2)=0 when x=-2 rarr y=0 i.e. the vertex ...

This means that anything less than five is an acceptable answer to this inequality. See below for directions on how to graph it: graph {x<5 [-10, 10, -5, 5]}