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Question 713419: Graph the quadric function. give vertex, axis or symmetry, domain, and range.
f(x)=-2x^2+20x-42
Answer by persian52(161)   (Show Source):
You can put this solution on YOUR website! █ Vertex █
f(x)=-2x^(2)+20x-42
To create a trinomial square on the left-hand side of the equation, add a value to both sides of the equation that is equal to the square of half the coefficient of x. In this problem, add (-5)^(2) to both sides of the equation.
y=-2(x^(2)-10x+25)-2(21)-(-2)(0+25)
Remove the 0 from the polynomial; adding or subtracting 0 does not change the value of the expression.
y=-2(x^(2)-10x+25)-2(21)-(-2)(25)
Factor the perfect trinomial square into (x-5)^(2).
y=-2((x-5)^(2))-2(21)-(-2)(25)
Factor the perfect trinomial square into (x-5)^(2).
y=-2(x-5)^(2)-2(21)-(-2)(25)
Multiply -2 by each term inside the parentheses.
y=-2(x-5)^(2)-42-(-2)(25)
Multiply -1 by each term inside the parentheses.
y=-2(x-5)^(2)-42+50
Add 50 to -42 to get 8.
y=-2(x-5)^(2)+8
This is the form of a paraboloa. Use this form to determine the values used to find vertex and x-y intercepts.
y=a(x-h)^(2)+k
The vertex of a parabola is (h,k).
Solution --> Vertex: (5,8)
█ Axis of Symmetry █
For any quadratic equation the axis of symmetry will be this:
x=-b/2a
f(x)=-2x^2+20x-42
b= 20, a= -2
Solution --> x= -20/2(-2)= 10
█ Domain █
f(x)=-2x^(2)+20x-42
The domain of the rational expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.
Solution --> All real numbers
█ Range █
f(x)=-2x^(2)+20x-42
Since x is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation.
-2x^(2)+20x-42=f(x)
To set the left-hand side of the equation equal to 0, move all the expressions to the left-hand side.
-2x^(2)+20x-f(x)-42=0
Divide each term in the equation by -1.
2x^(2)-20x+f(x)+42=0
Use the quadratic formula to find the solutions. In this case, the values are a=2, b=-20, and c=1f(x)+42.
x=(-b\~(b^(2)-4ac))/(2a) where ax^(2)+bx+c=0
Substitute in the values of a=2, b=-20, and c=1f(x)+42.
x=(-(-20)\~((-20)^(2)-4(2)(1f(x)+42)))/(2(2))
Multiply -1 by each term inside the parentheses.
x=(20\~((-20)^(2)-4(2)(1f(x)+42)))/(2(2))
Simplify the section inside the radical.
x=(20\2~(2(8-f(x))))/(2(2))
Simplify the denominator of the quadratic formula.
x=(20\2~(2(8-f(x))))/(4)
Simplify the expression to solve for the + portion of the \.
x=(10+~(2(8-f(x))))/(2)
Simplify the expression to solve for the - portion of the \.
x=(10-~(2(8-f(x))))/(2)
The final answer is the combination of both solutions.
x=(10+~(2(8-f(x))))/(2),(10-~(2(8-f(x))))/(2)
The domain of an expression is all real numbers except for the regions where the expression is undefined. This can occur where the denominator equals 0, a square root is less than 0, or a logarithm is less than or equal to 0. All of these are undefined and therefore are not part of the domain.
(2(8-f(x)))<0
The domain of the rational expression is all real numbers except where the expression is undefined.
f(x)<=8_(-I,8]
The domain of the inverse of f(x)=-2x^(2)+20x-42 is equal to the range of f(f(x))=((10+~(2(8-f(x)))))/(2).
Solution --> Range: f(x)<=8_(-I,8]
█ Graph █
http://s5.postimage.org/lblbh3llj/graph.png
I hope that helped.
Enjoy, and good luck studying.
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