Question

find the range of the quadratic function write your answer as an inequality using x or y as appropriate. or you may instead click on empty set or all realsas the answer

Answer 1

Explanation

The range of a function is a set composed of all of its output values. In this case this means all the possible y-values that the function can take. This is a cuadratic function which means that it has either a minimum or a maximum y-value that delimits its range. The sign of its leading coefficient is - which implies that this function has a maximum. This maximum is also known as the vertex of the function and if it's the point (h,k) then the range of this function can be described with this inequality:

$y\leq k$

So if we find the y-value of the vertex we find the range. In order to find it we could find its x-value first. If the function has two x-intercepts then the x-value of the vertex is the midvalue of these two. Then we should find the x-intercepts so we find the x-value of the vertex and with it its y-value.

For any given quadratic equation of the form ax²+bx+c=0 its x-intercepts are given by:

$r=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

In our case we have a=-3, b=30 and c=-72 so we get:

$\begin{gathered} r=\frac{-30\pm\sqrt{30^2-4\cdot(-3)\cdot(-72)}}{2\cdot(-3)}=\frac{-30\pm\sqrt{36}}{-6}=\frac{-30\pm6}{-6} \\ r=\frac{-30+6}{-6}=4\text{ and }r=\frac{-30-6}{-6}=6 \end{gathered}$

So the x-values of the x-intercepts are 4 and 6. Then the x-value of the vertex (h) of this function is:

$h=\frac{4+6}{2}=\frac{10}{2}=5$

Then the y-value of the vertex k is given by taking x=5 in the equation:

$k=-3h^2+30h-72=-3\cdot5^2+30\cdot5-72=3$Answer

Then the answer is:

$y\leq3$