Answer:
2: 15
3: 20
for the second chart
0: -2
2: 8
4: 18
6: 28
I hope this helped :))
i hope this is correct '~'
Answer: c) 3
<u>Step-by-step explanation:</u>
Plug in y = -9 and the given x-value to see if it makes a true statement.
a) -4.5 falls in the piecewise function x < -3 --> y = -x
y = -x
-9 = -(-4.5)
-9 = 4.5 FALSE
b) -3 falls in the piecewise function -3 ≤ x ≤ -2 --> y = 2x
y = 2x
-9 = 2(-3)
-9 = -6 FALSE
c) 3 falls in the piecewise function x > -2 --> y = -x²
y = -x²
-9 = -(-3)²
-9 = -9 TRUE
d) 9 falls in the piecewise function x > -2 --> y = -x²
y = -x²
-9 = -(9)²
-9 = -81 FALSE
Answer:
Step-by-step explanation:
f(x) = 2x - 3
f(-2) = 2(-2) -3 = -4 - 3 = -7
f(-1) = 2(-1) -3 = -2 - 3 = - 5
f(0) = 2(0) -3 = 0 - 3 = -3
f(1) = 2(1) -3 = 2 - 3 = -1
f(2) = 2(2) -3 = 4 - 3 = 1
f(3) = 2(3) -3 = 6 - 3 = 3
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.
