SOLUTION: find the vertex of the parobola f(x)=-4x^2+40x-93

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: find the vertex of the parobola f(x)=-4x^2+40x-93      Log On


   

Question 109691: find the vertex of the parobola f(x)=-4x^2+40x-93
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
To find the vertex, we need to know the axis of symmetry

To find the axis of symmetry, use this formula:

x=-b%2F%282a%29

From the equation y=-4x%5E2%2B40x-93 we can see that a=-4 and b=40

x=%28-40%29%2F%282%2A-4%29 Plug in b=40 and a=-4


x=%28-40%29%2F-8 Multiply 2 and -4 to get -8



x=5 Reduce


So the axis of symmetry is x=5


So the x-coordinate of the vertex is x=5. Lets plug this into the equation to find the y-coordinate of the vertex.


Lets evaluate f%285%29

f%28x%29=-4x%5E2%2B40x-93 Start with the given polynomial


f%285%29=-4%285%29%5E2%2B40%285%29-93 Plug in x=5


f%285%29=-4%2825%29%2B40%285%29-93 Raise 5 to the second power to get 25


f%285%29=-100%2B40%285%29-93 Multiply 4 by 25 to get 100


f%285%29=-100%2B200-93 Multiply 40 by 5 to get 200


f%285%29=7 Now combine like terms


So the vertex is (5,7)



Notice if we graph the equation y=-4x%5E2%2B40x-93 we get

and we can see that the vertex is (5,7)