Chapter 2.1, Problem 52E

To determine

To find: two quadratic functions, one that open upward and one that opens downward.

Answer to Problem 52E

One that opens upward, $x^2-10x$

One that open downward, $-x^2+10x$

Explanation of Solution

Given information: x-intercepts of quadratic function is given as follows-

  $(0,0),(10,0)$.

The x-intercepts of quadratic function are $(0,0),(10,0)$.

Then zeroes of quadratic function are $\text{0 and 10}$

Sum of zeroes = 0+10=10

Product of zeroes = $0\times 10=0$

Then quadratic equation whose zeroes $\text{0 and 10}$ is given by

  $x^2-x(\text{sum of zeroes)+(product of zeroes)=0}$

  $\begin{array}{l}\Rightarrow {\text{x}}^2-x(10)-0=0\\ \Rightarrow x^2-10x=0\end{array}$

Hence required quadratic function that open upward is $x^2-10x$ .

Here coefficient of $x^2$ is positive so it is open upward,

To find a quadratic function which is open downward multiply the equation by ‘−‘

  $\Rightarrow -x^2+10x=0$

Hence required quadratic function that open downward is $-x^2+10x$ .