Question

What is the range of `f(x)=25x^2-30x+16`?

Answer 1

`y>=1129/144`

Explanation:

The range of a function is the set of all the y-coordinates that are represented in the function. A good way to tell range can be simply through looking at a graph.
graph{25x^2-30x+16 [-15.19, 16.85, -2.13, 15.56]}
This is the graph of `f(x)=25x^2-30x+16`. It appears as if the range, or all the y-values that the graph "covers" starts at about `7` and continues on to `oo`.

We can determine the range algebraically. The vertex of the parabola is the lowest point of the function, so, if we can determine its y-value, we know where the range begins.

So, to figure out the location of the vertex, we can use the vertex formula for a parabola `(-(b)/(2a),f(-(b)/(2a)))`. The `a` and `b` come from the standard form of the parabola `ax^2+bx+c`, so, for `25x^2-30x+16`, `a=25` and `b=-30`.

First, we figure out the x-coordinate of the vertex by plugging in `a` and `b`.
`-(-25)/(2xx30)=25/60=5/12`

Then, to figure out the y-coordinate (this is the one that matters), we find `f(5/12)`, that is, plug `5/12` into the original equation.

We get:
`f(5/12)=25(5/12)^2-30(5/12)+16`
`=25(25/144)-30(60/144)+2304/144`
`=625/144-1800/144+2304/144`
`=color(blue)(1129/144`

Therefore, the coordinate of the vertex is `(5/12,1129/144)`.
This means that the function's lowest y-value is `1129/144`, so the range can be written as `y>=1129/144,{y|y>=1129/144}` or `[1129/144,oo)`.