# Graph the quadratic function. Give the vertex, axis of symmetry, domain, and range. f(x)=-2x^2+20x-40

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### 2 Answers

To find the vertex of a quadratic function, we need to complete the square of the function.

`f(x)=-2x^2+20x-40` factor the first two terms

`=-2(x^2-10x)-40` turn the brackets into a perfect square

`=-2(x^2-10x+25-25)-40` expand the last term in the brackets

`=-2(x^2-10x+25)+50-40` simplify the brackets

`=-2(x-5)^2+10`

The vertex is at `(5,10)` . Now the domain is always all real numbers for quadratic functions. Since the function opens down, then the range is `{y in R|y<=10}` . The axis of symmetry is at `x=5` which is the x-value of the vertex.

`f(x)=-2x^2+20x-40`

`x=-b/(2a)`

`a=-2` `b=20 ` ` c=-40`

`x=-20/(2xx-2) ` ` x= -20/-4 ` `x=5`

plug in the 5 into the equation

`-2(5)^2+20(5)-40`

`-2(25)+100-40`

`-50+100-40`

`-50+100=50-40=10`

`y=10`

vertex is `(5,10)`

the aos is always the x, so `aos=5`

domain is all real numbers

the range is `y<=10` the is because a is negative