# for the following function find the vertex(h,k), axis sym. domain f, range f , and the intercepts f(x)=-4x^2 +40x -105

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Given `f(x)=-4x^2+40x-105` :

Write in vertex form `f(x)=a(x-h)^2+k` by completing the square.

`f(x)=-4x^2+40x-105`

`=-4(x^2-10x)-105`

`=-4(x^2-10x+25-25)-105`

`=-4(x^2-10x+25)-5`

`=-4(x-5)^2-5`

(a) The vertex is at `(h,k)` or `(5,-5)`

(b) The domain is all real numbers since this is a polynomial.

(c) The range is `y<=-5` or `(-oo,-5]`

(d) The y-intercept is found by setting x=0; the y-intercept is -105

(e) There are no x-intercepts. The graph is a parabola (it is a quadratic function) opening down (since a=-5) and the vertex is beneath the x-axis, thus there are no x-intercepts.

** If you use the quadratic formula you get the discriminant `b^2-4ac=40^2-4(-4)(-105)=-80` . Since the discriminant is less than 0 there are no real solutions thus no intercepts.

The graph: