The equations of the functions are y = -4(x + 1)^2 + 2, y = 2(x - 2)^2 + 1 and y = -(x - 1)^2 - 2
<h3>How to determine the functions?</h3>
A quadratic function is represented as:
y = a(x - h)^2 + k
<u>Question #6</u>
The vertex of the graph is
(h, k) = (-1, 2)
So, we have:
y = a(x + 1)^2 + 2
The graph pass through the f(0) = -2
So, we have:
-2 = a(0 + 1)^2 + 2
Evaluate the like terms
a = -4
Substitute a = -4 in y = a(x + 1)^2 + 2
y = -4(x + 1)^2 + 2
<u>Question #7</u>
The vertex of the graph is
(h, k) = (2, 1)
So, we have:
y = a(x - 2)^2 + 1
The graph pass through (1, 3)
So, we have:
3 = a(1 - 2)^2 + 1
Evaluate the like terms
a = 2
Substitute a = 2 in y = a(x - 2)^2 + 1
y = 2(x - 2)^2 + 1
<u>Question #8</u>
The vertex of the graph is
(h, k) = (1, -2)
So, we have:
y = a(x - 1)^2 - 2
The graph pass through (0, -3)
So, we have:
-3 = a(0 - 1)^2 - 2
Evaluate the like terms
a = -1
Substitute a = -1 in y = a(x - 1)^2 - 2
y = -(x - 1)^2 - 2
Hence, the equations of the functions are y = -4(x + 1)^2 + 2, y = 2(x - 2)^2 + 1 and y = -(x - 1)^2 - 2
Read more about parabola at:
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Answer................:8x^2-2x-1
This graph is composed of four straight line segments. You'll need to determine the slope, y-intercept and domain for each of them. Look at the first segment, the one on the extreme left. Verify yourself that the slope of this line segment is 1 and that the y-intercept would be 0 if you were to extend this segment all the way to the y-axis. Thus, the rule (formula, equation) for this line segment would be f(x)=1x+0, or just f(x)=x, for (-3,-1). Use a similar approach to write rules for the remaining three line segments.
Present your answer like this:
x, (-3,-1)
f(x) = -1, (-1,0)
one more here
one more here
Answer:
The answer depends on what the problem really is.
f(1/2) = 8
or
f(1/2) = 4
See below.
Step-by-step explanation:
This is the problem as it was posted:
y = f(x) = 16x
Let x = 1/2
y = f(1/2) = 16(1/2) = 8
Perhaps the problem is this:
If x is an exponent of 16 and not a factor, then the answer is:
