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

Nov 15, 2015

$y \ge \frac{1129}{144}$

#### 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 $f \left(x\right) = 25 {x}^{2} - 30 x + 16$. It appears as if the range, or all the y-values that the graph "covers" starts at about $7$ and continues on to $\infty$.

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 $\left(- \frac{b}{2 a} , f \left(- \frac{b}{2 a}\right)\right)$. The $a$ and $b$ come from the standard form of the parabola $a {x}^{2} + b x + c$, so, for $25 {x}^{2} - 30 x + 16$, $a = 25$ and $b = - 30$.

First, we figure out the x-coordinate of the vertex by plugging in $a$ and $b$.
$- \frac{- 25}{2 \times 30} = \frac{25}{60} = \frac{5}{12}$

Then, to figure out the y-coordinate (this is the one that matters), we find $f \left(\frac{5}{12}\right)$, that is, plug $\frac{5}{12}$ into the original equation.

We get:
$f \left(\frac{5}{12}\right) = 25 {\left(\frac{5}{12}\right)}^{2} - 30 \left(\frac{5}{12}\right) + 16$
$= 25 \left(\frac{25}{144}\right) - 30 \left(\frac{60}{144}\right) + \frac{2304}{144}$
$= \frac{625}{144} - \frac{1800}{144} + \frac{2304}{144}$
=color(blue)(1129/144

Therefore, the coordinate of the vertex is $\left(\frac{5}{12} , \frac{1129}{144}\right)$.
This means that the function's lowest y-value is $\frac{1129}{144}$, so the range can be written as $y \ge \frac{1129}{144} , \left\{y | y \ge \frac{1129}{144}\right\}$ or $\left[\frac{1129}{144} , \infty\right)$.