SOLUTION: 1. From a square piece of cardboard with wisth x inches, a square of width x-3 inches is removed from the center. Write the area of the remaining piece as a function of x. 2. If

Algebra ->  Points-lines-and-rays -> SOLUTION: 1. From a square piece of cardboard with wisth x inches, a square of width x-3 inches is removed from the center. Write the area of the remaining piece as a function of x. 2. If      Log On


   



Question 67835: 1. From a square piece of cardboard with wisth x inches, a square of width x-3 inches is removed from the center. Write the area of the remaining piece as a function of x.
2. If P(4, -5) is a point on the graph of the function y=f(x), find the corresponding point on the graph of y=2f(x-6).
3. Explain how the graph of y-5=(x-3)2 power can be obtained from the graph of y=x2.
4. Determine the vertex of y=x2-8x+22.
5. Find the maximum value of y=-x2 + 6x.
6. Given f(x)=5x+7 and g(x)=x2+7, find (g o f)(x).

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
1. From a square piece of cardboard with wisth x inches, a square of width x-3 inches is removed from the center. Write the area of the remaining piece as a function of x.
Remaining area = original area - area cut out
A(x) = x^2 - (x-3)^2
A(x) = x^2-[x^2-6x+9]
A(x) = 6x-9
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2. If P(4, -5) is a point on the graph of the function y=f(x), find the corresponding point on the graph of y=2f(x-6).
(x-6) moves the point 6 to the right to (10,-5)
2 doubles the y value so you end up at (10,-10)
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3. Explain how the graph of y-5=(x-3)^2 power can be obtained from the graph of y=x^2.
(x-3) moves the parabola 3 to the right.
5 then moves the parabola 5 up.
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4. Determine the vertex of y=x^2-8x+22.
The x-value is at -b/2a = 8/2=4
Then f(4)=16-32+22 = 6
Vertex at (4,6)
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5. Find the maximum value of y=-x^2 + 6x.
max occurs at the vertex where x=-b/2a = -6/-2=3
f(3)=-9+18=9
Max at (3,9)
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6. Given f(x)=5x+7 and g(x)=x^2+7, find (g o f)(x).
g[5x+7]=(5x+7)^2+7 = 25x^2+70x+56
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Cheers,
Stan H.