the following table shows the revenue for a company generates based on the increases in the price of the product. What is the y-value of the Vertex of the parabola that models the date?
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Answer:
The y-value of the Vertex of the parabola that models the data is 1125.
Step-by-step explanation:
Let the function of parabola is
[tex]f(x)=ax^2+bx+c[/tex]
From the given that it is noticed that the parabolic function passing through the points (1,1045), (3,1105) and (5,1125). It means the function must be satisfied by these points.
[tex]1045=a(1)^2+b(1)+c[/tex]
[tex]1045=a+b+c[/tex] ....(1)
[tex]1105=a(3)^2+b(3)+c[/tex]
[tex]1105=9a+3b+c[/tex] ....(2)
[tex]1125=a(5)^2+b(5)+c[/tex]
[tex]1125=25a+5b+c[/tex] ....(3)
On solving (1), (2) and (3) we get,
[tex]a=-5[/tex]
[tex]b=50[/tex]
[tex]c=1000[/tex]
Therefore the equation of parabola is
[tex]f(x)=-5x^2+50x+1000[/tex]
The vertex of the parabola is
[tex](\frac{-b}{2a},f(\frac{-b}{2a}))[/tex]
[tex]\frac{-b}{2a}=-\frac{50}{2(-5)}=5[/tex]
[tex]f(5)=1125[/tex]
Therefore the vertex is (5,1125) and y-value of the Vertex of the parabola that models the data is 1125.
The vertexes of the parabola are, (5, 1125).
Explanation
The table given to us in the problem are the data points that will lie on the parabola, therefore,
Point 1 = (1, 1045)
Point 2 = (3, 1105)
Point 3 = (5, 1125)
Point 4 = (3, 1105)
Point 5 = (1, 1045)
We know that the equation of a parabola is given as,
[tex]y = ax^2 +bx+c[/tex]
For point 1,
Point 1 = (1, 1045)
Substituting the value in the equation of a parabola,
[tex]1045 = a(1)^2 +(1)b+c\\\\1045 = a+b+c[/tex]..... equation 1,
For point 2,
Point 2 = (3, 1105)
Substituting the value in the equation of a parabola,
[tex]1105 = a(3)^2 +(3)b+c\\\\1105= 9a+3b+c[/tex]..... equation 2,
For point 3,
Point 3 = (5, 1125)
Substituting the value in the equation of a parabola,
[tex]1125= a(5)^2 +(5)b+c\\\\1125= 25a+5b+c[/tex]..... equation 3,
Solving the three equations we get,
a = -5,
b = 50,
c = 1000
Substitute the values in the equation of a parabola,
[tex]y=f(x) = -5x^2 +50x +1000[/tex]
To find the vertex of a parabolic equation we bring the equation into the form,
[tex]y = a(x-h)+k\\[/tex] , where h and k are the vertexes of the parabola.
Vertex of the Parabola,
[tex]y=f(x) = -5x^2 +50x +1000\\\\y = -5x^2 +50x +1000\\\\y =-5(x^2 -10x)+1000\\\\y =-5(x^2 -10x+25-25)+1000\\\\y =-5(x^2 -10x+25)+ (-5\times -25)+1000\\\\y =-5(x^2 -10x +25)+125+1000\\\\y =-5(x^2 -5)^2+1125[/tex]
Comparing it to the equation, [tex]y = a(x-h)+k\\[/tex],
the vertexes of the parabola are,
(5, 1125)
Learn more about the Equation of a Parabola:
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