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:
0.2
Step-by-step explanation:
Well, first put the function in slop-intercept form (y = mx + b). 2x + y = 12, y - 2x = 12 - 2x, y = -2x + 12. So y = -2x + 12 is the slope-intercept form. Then graph the function to solve for all possible solutions to the function. In y = mx + b, m is the slope and b is the y intercept. In y = -2x + 12, m equals -2, so the slope is -2, and b equals 12, so 12 is the y intercept. Make a point on the y intercept at y = 12. The slope is -2, or -2/1. The graph will be sloping downward. From the point at y = 12, move down two units and to the right one unit and make another point. Then from this new point move down two units and to the right one unit and make another point, and so on. Then to extend the graph in the other direction, start at the point y = 12 and move two units up and one unit to the right and make a point there. Then from this new point move two units up and one unit to the right and make another point there, and so on.
