Answer:1928
Step-by-step explanation:488* 6
I used lattice
Answer:
I act answer it sorry for that
Y1 is the simplest parabola. Its vertex is at (0,0) and it passes thru (2,4). This is enough info to conclude that y1 = x^2.
y4, the lower red graph, is a bit more of a challenge. We can easily identify its vertex, which is (-4,0), and several points on the grah, such as (2,-3).
Let's try this: assume that the general equation for a parabola is
y-k = a(x-h)^2, where (h,k) is the vertex. Subst. the known values,
-3-(-4) = a(2-0)^2. Then 1 = a(2)^2, or 1 = 4a, or a = 1/4.
The equation of parabola y4 is y+4 = (1/4)x^2
Or you could elim. the fraction and write the eqn as 4y+16=x^2, or
4y = x^2-16, or y = (1/4)x - 4. Take your pick! Hope this helps you find "a" for the other parabolas.
Step-by-step explanation:
Set up the composite result function.
f(g(x))f(g(x))
Evaluate f(g(x))f(g(x)) by substituting in the value of gg into ff.
f(x2)=2(x2)−4f(x2)=2(x2)-4
Multiply 22 by x2x2.
f(x2)=2x2−4f(x2)=2x2-4
